RD Sharma solution class 8 chapter 26 Data Handling IV(Probability) Exercise 26.1

Exercise 26.1

Page-26.14

Question 1:

The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?

Answer 1:

$$\begin{gathered} \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{raining} \mathrm{tomorrow}. \\ \mathrm{The} \mathrm{probability} \mathrm{that} \mathrm{it} \mathrm{will} \mathrm{rain} \mathrm{tomorrow}, \mathrm{P}\left(\mathrm{A}\right), \mathrm{is} 0.85. \\ \mathrm{Since} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{raining} \mathrm{tomorrow} \mathrm{and} \mathrm{not} \mathrm{raining} \mathrm{tomorrow} \mathrm{are} \mathrm{complementary} \mathrm{to} \mathrm{each} \mathrm{other}, \mathrm{the} \mathrm{probability} \mathrm{of} \mathrm{not} \mathrm{raining} \mathrm{tomorrow} \mathrm{is}: \\ \mathrm{P}\left(\overline{\mathrm{A}}\right) = 1 - \mathrm{P}\left(\mathrm{A}\right) = 1 - 0.85 = 0.15 \end{gathered}$$

Question 2:

A die is thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3

Answer 2:

$$\begin{gathered} \mathrm{When} \mathrm{a} \mathrm{die} \mathrm{is} \mathrm{thrown}, \mathrm{the} \mathrm{possible} \mathrm{outcomes} \mathrm{are} 1, 2, 3, 4, 5 \mathrm{and} 6. \\ \mathrm{Thus}, \mathrm{the} \mathrm{sample} \mathrm{space} \mathrm{will} \mathrm{be} \mathrm{as} \mathrm{follows}: \\ \mathrm{S}=\left\{1,2,3,4,5,6\right\} \\ \left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{prime} \mathrm{number}. \\ \mathrm{There} \mathrm{are} 3 \mathrm{prime} \mathrm{numbers} (2,3 \mathrm{and} 5) \mathrm{in} \mathrm{the} \mathrm{sample} \mathrm{space}. \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 3. \\ \mathrm{Hence}, \mathrm{the} \mathrm{probability} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{prime} \mathrm{number} \mathrm{is} \mathrm{as} \mathrm{follows}: \\ \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{3}{6}=\frac{1}{2} \\ \left(\mathrm{ii}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{two} \mathrm{or} \mathrm{four}. \\ \mathrm{Two} \mathrm{or} \mathrm{four} \mathrm{occurs} \mathrm{once} \mathrm{in} \mathrm{a} \mathrm{single} \mathrm{roll}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 2. \\ \mathrm{Hence}, \mathrm{the} \mathrm{probability} \mathrm{of} \mathrm{getting} 2 \mathrm{or} 4 \mathrm{is} \mathrm{as} \mathrm{follows}: \\ \mathrm{P}\left(\mathrm{A}\right)=\frac{2}{6}=\frac{1}{3} \\ \left(\mathrm{iii}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{multiples} \mathrm{of} 2 \mathrm{or} 3. \\ \mathrm{Here}, \mathrm{the} \mathrm{multiples} \mathrm{of} 2 \mathrm{are} 2, 4, 6 \mathrm{and} \mathrm{the} \mathrm{multiples} \mathrm{of} 3 \mathrm{are} 3 \mathrm{and} 6. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{favourable} \mathrm{outcomes} \mathrm{are} 2, 3, 4 \mathrm{and} 6. \\ \mathrm{Hence}, \mathrm{the} \mathrm{probability} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{multiple} \mathrm{of} 2 \mathrm{or} 3 \mathrm{is} \mathrm{as} \mathrm{follows}: \\ \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favorable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{4}{6}=\frac{2}{3} \end{gathered}$$

Page-26.15

Question 3:

In a simultaneous throw of a  pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.

Answer 3:

$$\begin{gathered} \mathrm{When} \mathrm{a} \mathrm{pair} \mathrm{of} \mathrm{dice} \mathrm{is} \mathrm{thrown} \mathrm{simultaneously}, \mathrm{the} \mathrm{sample} \mathrm{space} \mathrm{will} \mathrm{be} \mathrm{as} \mathrm{follows}: \\ \mathrm{S}=\left\{\left(1,1\right), \left(1,2\right), \left(1,3\right), \left(1,4\right),\cdots \left(6,5\right), \left(6,6\right)\right\} \\ \mathrm{Hence}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{outcomes} \mathrm{is} 36. \\ \left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} 8. \\ \mathrm{Now}, \mathrm{the} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} 8 \mathrm{are} \left(2,6\right), \left(3,5\right), \left(4,4\right), \left(5,3\right) \mathrm{and} \left(6,2\right). \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 5. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{5}{36} \\ \left(\mathrm{ii}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{doublets} \mathrm{in} \mathrm{the} \mathrm{sample} \mathrm{space}. \\ \mathrm{The} \mathrm{doublets} \mathrm{in} \mathrm{the} \mathrm{sample} \mathrm{space} \mathrm{are} \left(1,1\right), \left(2,2\right), \left(3,3\right), \left(4,4\right), \left(5,5\right) \mathrm{and} \left(6,6\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 6. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{6}{36}=\frac{1}{6} \\ \left(\mathrm{iii}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{doublets} \mathrm{of} \mathrm{prime} \mathrm{numbers} \mathrm{in} \mathrm{the} \mathrm{sample} \mathrm{space}. \\ \mathrm{The} \mathrm{doublets} \mathrm{of} \mathrm{prime} \mathrm{numbers} \mathrm{in} \mathrm{the} \mathrm{sample} \mathrm{space} \mathrm{are} \left(2,2\right), \left(3,3\right) \mathrm{and} \left(5,5\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 3. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{3}{36}=\frac{1}{12} \\ \left(\mathrm{iv}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{doublets} \mathrm{of} \mathrm{odd} \mathrm{numbers} \mathrm{in} \mathrm{the} \mathrm{sample} \mathrm{space}. \\ \mathrm{The} \mathrm{doublets} \mathrm{of} \mathrm{odd} \mathrm{numbers} \mathrm{in} \mathrm{the} \mathrm{sample} \mathrm{space} \mathrm{are} \left(1,1\right), \left(3,3\right) \mathrm{and} \left(5,5\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 3. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{3}{36}=\frac{1}{12} \\ \left(\mathrm{v}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} \mathrm{greater} \mathrm{than} 9. \\ \mathrm{The} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} \mathrm{greater} \mathrm{than} 9 \mathrm{are} \left(4,6\right),\left(5,5\right), \left(5,6\right),\left(6,4\right),\left(6,5\right) \mathrm{and} \left(6,6\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favorable} \mathrm{outcomes} \mathrm{is} 6. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{6}{36}=\frac{1}{6} \\ \left(\mathrm{vi}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{pairs} \mathrm{who} \mathrm{has} \mathrm{even} \mathrm{numbers} \mathrm{on} \mathrm{first} \mathrm{in} \mathrm{the} \mathrm{sample} \mathrm{space}. \\ \mathrm{The} \mathrm{pairs} \mathrm{who} \mathrm{has} \mathrm{even} \mathrm{numbers} \mathrm{on} \mathrm{first} \mathrm{are}: \left(2,1\right), \left(2,2\right),\dots \left(2,6\right), \left(4,1\right),\cdots ,\left(4,6\right), \left(6,1\right),\cdots \left(6,6\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 18. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{18}{36}=\frac{1}{3} \\ \left(\mathrm{vii}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{pairs} \mathrm{with} \mathrm{an} \mathrm{even} \mathrm{number} \mathrm{on} \mathrm{one} \mathrm{die} \mathrm{and} \mathrm{a} \mathrm{multiple} \mathrm{of} 3 \mathrm{on} \mathrm{the} \mathrm{other}. \\ \mathrm{The} \mathrm{pairs} \mathrm{with} \mathrm{an} \mathrm{even} \mathrm{number} \mathrm{on} \mathrm{one} \mathrm{die} \mathrm{and} \mathrm{a} \mathrm{multiple} \mathrm{of} 3 \mathrm{on} \mathrm{the} \mathrm{other} \mathrm{are} \left(2,3\right), \left(2,6\right), \left(4,3\right), \left(4,6\right), \left(6,3\right) \mathrm{and} \left(6,6\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 6. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{6}{36}=\frac{1}{6} \\ \left(\mathrm{viii}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} 9 \mathrm{or} 11. \\ \mathrm{The} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} 9 \mathrm{are} \left(3,6\right), \left(4,5\right), \left(5,4\right) \mathrm{and} \left(6,3\right). \\ \mathrm{And}, \mathrm{the} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} 11 \mathrm{are} \left(5,6\right) \mathrm{and} \left(6,5\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 6. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{6}{36}=\frac{1}{6} \\ \therefore \mathrm{P}\left(\mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{pairs} \mathrm{with} \mathrm{neither} 9 \mathrm{nor} 11\right) = 1 - \mathrm{P}\left(\mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{pairs} \mathrm{having} 9 \mathrm{or} 11\right) = 1 - \frac{1}{6} = \frac{5}{6} \\ \left(\mathrm{ix}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} \mathrm{less} \mathrm{than} 6. \\ \mathrm{The} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} \mathrm{less} \mathrm{than} 6 \mathrm{are} \left(1,1\right), \left(1,2\right), \left(1,3\right), \left(1,4\right), \left(2,1\right), \left(2,2\right), \left(2,3\right), \left(3,1\right), \left(3,2\right) \mathrm{and} \left(4,1\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 10. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{10}{36}=\frac{5}{18} \\ \left(\mathrm{x}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} \mathrm{less} \mathrm{than} 7. \\ \mathrm{The} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} \mathrm{less} \mathrm{than} 7 \mathrm{are} \left(1,1\right), \left(1,2\right), \left(1,3\right), \left(1,4\right), \left(1,5\right), \left(2,1\right), \left(2,2\right), \left(2,3\right), \left(2,4\right), \left(3,1\right), \left(3,2\right), \left(3,3\right), \left(4,1\right), \left(4,2\right) \mathrm{and} \left(5,1\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 15. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{15}{36}=\frac{5}{12} \\ \left(\mathrm{xi}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} \mathrm{more} \mathrm{than} 7. \\ \mathrm{The} \mathrm{pairs} \mathrm{whose} \mathrm{sum} \mathrm{is} \mathrm{more} \mathrm{than} 7 \mathrm{are} \left(2,6\right), \left(3,5\right), \left(3,6\right), \left(4,4\right), \left(4,5\right), \left(4,6\right), \left(5,3\right), \left(5,4\right), \left(5,5\right), \left(5,6\right), \left(6,2\right), \left(6,3\right), \left(6,4\right), \left(6,5\right) \mathrm{and} \left(6,6\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 15. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{15}{36}=\frac{5}{12} \\ \left(\mathrm{xii}\right) \mathrm{Incomplete} \mathrm{question} \\ \left(\mathrm{xiii}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{pairs} \mathrm{that} \mathrm{has} \mathrm{the} \mathrm{number} 5. \\ \mathrm{The} \mathrm{pairs} \mathrm{that} \mathrm{has} \mathrm{the} \mathrm{number} 5 \mathrm{are} \left(1,5\right), \left(2,5\right), \left(3,5\right), \left(4,5\right), \left(5,1\right), \left(5,2\right), \left(5,3\right), \left(5,4\right), \left(5,5\right), \left(5,6\right), \left(6,1\right), \left(6,2\right), \left(6,3\right), \left(6,4\right) \mathrm{and} \left(6,6\right). \\ \mathrm{Hence}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 11. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{11}{36} \\ \therefore \mathrm{P}\left(\overline{\mathrm{A}}\right)=1-\mathrm{P}\left(\mathrm{A}\right)=1-\frac{11}{36}=\frac{25}{36} \end{gathered}$$

Question 4:

Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail
(iv) no tails

Answer 4:

$$\begin{gathered} \mathrm{When} 3 \mathrm{coins} \mathrm{are} \mathrm{tossed} \mathrm{together}, \mathrm{the} \mathrm{outcomes} \mathrm{are} \mathrm{as} \mathrm{follows}: \\ \mathrm{S}=\left\{\left(\mathrm{h},\mathrm{h},\mathrm{h}\right), \left(\mathrm{h},\mathrm{h},\mathrm{t}\right), \left(\mathrm{h},\mathrm{t},\mathrm{h}\right), \left(\mathrm{h},\mathrm{t},\mathrm{t}\right), \left(\mathrm{t},\mathrm{h},\mathrm{h}\right), \left(\mathrm{t},\mathrm{h},\mathrm{t}\right), \left(\mathrm{t},\mathrm{t},\mathrm{h}\right), \left(\mathrm{t},\mathrm{t},\mathrm{t}\right)\right\} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{outcomes} \mathrm{is} 8. \\ \left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{triplets} \mathrm{having} \mathrm{exactly} 2 \mathrm{heads}. \\ \mathrm{Triplets} \mathrm{having} \mathrm{exactly} 2 \mathrm{heads}: \left(\mathrm{h},\mathrm{h},\mathrm{t}\right), \left(\mathrm{h},\mathrm{t},\mathrm{h}\right), \left(\mathrm{t},\mathrm{h},\mathrm{h}\right) \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 3. \\ \mathrm{P}\left(A\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{3}{8} \\ \left(\mathrm{ii}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{triplets} \mathrm{having} \mathrm{at} \mathrm{least} 2 \mathrm{heads}. \\ \mathrm{Triplets} \mathrm{having} \mathrm{at} \mathrm{least} 2 \mathrm{heads}: \left(\mathrm{h},\mathrm{h},\mathrm{t}\right), \left(\mathrm{h},\mathrm{t},\mathrm{h}\right), \left(\mathrm{t},\mathrm{h},\mathrm{h}\right), \left(\mathrm{h},\mathrm{h},\mathrm{h}\right) \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 4. \\ \mathrm{P}\left(A\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{4}{8}=\frac{1}{2} \\ \left(\mathrm{iii}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{triplets} \mathrm{having} \mathrm{at} \mathrm{least} \mathrm{one} \mathrm{head} \mathrm{and} \mathrm{one} \mathrm{tail}. \\ \mathrm{Triplets} \mathrm{having} \mathrm{at} \mathrm{least} \mathrm{one} \mathrm{head} \mathrm{and} \mathrm{one} \mathrm{tail}: \left(\mathrm{h},\mathrm{h},\mathrm{t}\right), \left(\mathrm{h},\mathrm{t},\mathrm{h}\right), \left(\mathrm{t},\mathrm{h},\mathrm{h}\right), \left(\mathrm{h},\mathrm{h},\mathrm{t}\right), \left(\mathrm{t},\mathrm{t},\mathrm{h}\right), \left(\mathrm{t},\mathrm{h},\mathrm{t}\right) \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 6. \\ \mathrm{P}\left(A\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{6}{8}=\frac{3}{4} \\ \left(\mathrm{iv}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{triplets} \mathrm{having} \mathrm{no} \mathrm{tail}. \\ \mathrm{Triplets} \mathrm{having} \mathrm{no} \mathrm{tail}: \left(\mathrm{h},\mathrm{h},\mathrm{h}\right) \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 1. \\ \mathrm{P}\left(A\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{1}{8} \end{gathered}$$

Question 5:

A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) neither a red card nor a queen.
(ix) other than an ace
(x) a ten
(xi) a spade
(xii) a black card
(xiii) the seven of clubs
(xiv) jack
(xv) the ace of spades
(xvi) a queen
(xvii) a heart
(xviii) a red card

Answer 5:

(i) There are two black kings, spade and clover. Hence, the probability that the drawn card is a black king is: 2/52 = 1/26
(ii) There are 26 black cards and 4 kings, but two kings are already black. Hence, we only need to count the red kings. Thus, the probability is: (26+2)/52 = 7/13
(iii) This question is exactly the same as part (i). Hence, the probability is: 2/52 = 1/26
(iv) There are 4 jacks, 4 queens and 4 kings in a deck. Hence, the probability of drawing either of them is: (4+4+4)/52 = 3/13
(v) This means that we have to leave the hearts and the kings out. There are 13 hearts and 3 kings (other than that of hearts). Hence, the probability of drawing neither a heart nor a king is: (52-13-3)/52 = 9/13
(vi) There are 13 spades and 3 aces (other than that of spades). Hence the probability is: (13+3)/52 = 4/13
(vii) This means that we have to leave the aces and the kings out. There are 4 aces and 4 kings. Hence, the probability of drawing neither an ace nor a king is: (52$-$4$-$4)/52 = 11/13.
(viii) This means that we have to leave the red cards and the queens out. There are 26 red cards and 2 queens (only black queens are counted since the reds are already counted among the red cards). Hence, the probability of drawing neither a red card nor a queen is: (52-26-2)/52 = 6/13
(ix) It means that we have to leave out the aces. Since there are 4 aces, then the probability is (52$-$4)/52 = 12/13
(x) Since there are four 10s, the probability is: 4/52 = 1/13
(xi) Since there are 13 spades, the probability is: 13/52 = 1/4
(xii) Since there are 26 black cards, the probability is: 26/52 = 1/2
(xiii) There is only one card named seven of clubs. Hence, the probability is 1/52.
(xiv) Since there are 4 jacks, the probability is: 4/52 = 1/13
(xv) There is only 1 card named ace of spade. Hence, the probability is 1/52.
(xvi) Since there are 4 queens, the probability is: 4/52 = 1/13
(xvii) Since there are 13 hearts, the probability is: 13/52 = 1/4
(xviii) Since there are 26 red cards, the probability is 26/52 = 1/2

Question 6:

An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.

Answer 6:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{red} \mathrm{balls} = 10 \\ \mathrm{Number} \mathrm{of} \mathrm{white} \mathrm{balls} = 8 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{balls} \mathrm{in} \mathrm{the} \mathrm{urn} = 10 + 8 = 18 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{cases} \mathrm{is} 18 \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{cases} \mathrm{is} 8. \\ \therefore \mathrm{P}\left(\mathrm{The} \mathrm{ball} \mathrm{drawn} \mathrm{is} \mathrm{white}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{cases}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{cases}}=\frac{8}{18}=\frac{4}{9} \end{gathered}$$

Question 7:

A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) white?
(ii) red?
(iii) black?
(iv) not red?

Answer 7:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{red} \mathrm{balls} = 3 \\ \mathrm{Number} \mathrm{of} \mathrm{black} \mathrm{balls} = 5 \\ \mathrm{Number} \mathrm{of} \mathrm{white} \mathrm{balls} = 4 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{balls} = 3 + 5 + 4 = 12 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{cases} \mathrm{is} 12. \\ \left(\mathrm{i}\right) \\ \mathrm{Since} \mathrm{there} \mathrm{are} 4 \mathrm{white} \mathrm{balls}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 4. \\ \mathrm{P}\left(\mathrm{a} \mathrm{white} \mathrm{ball}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{cases}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{cases}}=\frac{4}{12}=\frac{1}{3} \\ \left(\mathrm{ii}\right) \\ \mathrm{Since} \mathrm{there} \mathrm{are} 3 \mathrm{red} \mathrm{balls}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 3. \\ \mathrm{P}\left(\mathrm{a} \mathrm{red} \mathrm{ball}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{cases}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{cases}}=\frac{3}{12}=\frac{1}{4} \\ \left(\mathrm{iii}\right) \\ \mathrm{Since} \mathrm{there} \mathrm{are} 5 \mathrm{black} \mathrm{balls}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 5. \\ \mathrm{P}\left(\mathrm{a} \mathrm{black} \mathrm{ball}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{cases}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{cases}}=\frac{5}{12} \\ \left(\mathrm{iv}\right) \\ \mathrm{P}\left(\mathrm{not} \mathrm{a} \mathrm{red} \mathrm{ball}\right)=1-\mathrm{P}\left(\mathrm{a} \mathrm{red} \mathrm{ball}\right)=1-\frac{1}{4}=\frac{3}{4} \end{gathered}$$

Question 8:

What is the probability that a number selected from the numbers 1, 2, 3, ..., 15 is a multiple of 4?

Answer 8:

$$\begin{gathered} \mathrm{There} \mathrm{are} 15 \mathrm{numbers} \mathrm{from} 1, 2,\cdots ,15. \\ \mathrm{Hence}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{cases} \mathrm{is} 15. \\ \mathrm{Again}, \mathrm{the} \mathrm{multiples} \mathrm{of} 4 \mathrm{are} 4, 8 \mathrm{and} 12. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{cases} \mathrm{is} 3. \\ \therefore \mathrm{P}\left(\mathrm{the} \mathrm{number} \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 4\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{cases}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{cases}} =\frac{3}{15}=\frac{1}{5} \end{gathered}$$

Question 9:

A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?

Answer 9:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{red} \mathrm{balls} = 6 \\ \mathrm{Number} \mathrm{of} \mathrm{black} \mathrm{balls} = 8 \\ \mathrm{Number} \mathrm{of} \mathrm{white} \mathrm{balls} = 4 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{balls} = 6 + 8 + 4 = 18 \\ \therefore \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{cases} = 18 \\ \mathrm{Again}, \mathrm{number} \mathrm{of} \mathrm{balls} \mathrm{that} \mathrm{are} \mathrm{not} \mathrm{black} = 18 - 8 = 10 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{cases} \mathrm{is} 10. \\ \therefore \mathrm{P}\left(\mathrm{the} \mathrm{drawn} \mathrm{ball} \mathrm{is} \mathrm{not} \mathrm{black}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{cases}}{\mathrm{Total} \mathrm{number} \mathrm{cases}}=\frac{10}{18}=\frac{5}{9} \end{gathered}$$

Question 10:

A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?

Answer 10:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{white} \mathrm{balls} = 5 \\ \mathrm{Number} \mathrm{of} \mathrm{red} \mathrm{balls} = 7 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{balls} = 5 + 7 = 12 \\ \therefore \mathrm{The} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{cases} = 12 \\ \mathrm{Again}, \mathrm{there} \mathrm{are} 5 \mathrm{white} \mathrm{balls}. \mathrm{Therefore}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 5. \\ \therefore \mathrm{P}\left(\mathrm{the} \mathrm{drawn} \mathrm{ball} \mathrm{is} \mathrm{white}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{cases}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{cases} }=\frac{5}{12} \end{gathered}$$

Question 11:

A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
(i) white
(ii) red
(iii) not black
(iv) red or white

Answer 11:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{red} \mathrm{balls} = 4 \\ \mathrm{Number} \mathrm{of} \mathrm{black} \mathrm{balls} = 5 \\ \mathrm{Number} \mathrm{of} \mathrm{white} \mathrm{balls} = 6 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{balls} \mathrm{in} \mathrm{the} \mathrm{bag} = 4 + 5 + 6 = 15 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{cases} \mathrm{is} 15. \\ \left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{denote} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{white} \mathrm{ball}. \\ \mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}, \mathrm{i}.\mathrm{e}. \mathrm{number} \mathrm{of} \mathrm{white} \mathrm{balls} = 6 \\ \mathrm{P}\left(\mathrm{A}\right) =\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{6}{15}=\frac{2}{5} \\ \left(\mathrm{ii}\right) \\ \mathrm{Let} \mathrm{B} \mathrm{denote} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{red} \mathrm{ball}. \\ \mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}, \mathrm{i}.\mathrm{e}. \mathrm{number} \mathrm{of} \mathrm{red} \mathrm{balls} = 4 \\ \mathrm{P}\left(\mathrm{B}\right) =\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{4}{15} \\ \left(\mathrm{iii}\right) \\ \mathrm{Let} \mathrm{C} \mathrm{denote} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{black} \mathrm{ball}. \\ \mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}, \mathrm{i}.\mathrm{e}. \mathrm{number} \mathrm{of} \mathrm{black} \mathrm{balls} = 5 \\ \mathrm{P}\left(\mathrm{C}\right) =\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{5}{15}=\frac{1}{3} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{probabilty} \mathrm{of} \mathrm{not} \mathrm{getting} \mathrm{a} \mathrm{black} \mathrm{ball} \mathrm{is} \mathrm{as} \mathrm{follows}: \\ \mathrm{P}\left(\overline{\mathrm{C}}\right) = 1 - \mathrm{P}\left(\mathrm{C}\right) = 1 - \frac{1}{3} = \frac{2}{3} \\ \left(\mathrm{iv}\right) \\ \mathrm{Let} \mathrm{D} \mathrm{denote} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{red} \mathrm{or} \mathrm{a} \mathrm{white} \mathrm{ball}. \\ \mathrm{P}\left(\mathrm{D}\right) =\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{4 + 6}{15}=\frac{10}{15}=\frac{2}{5} \end{gathered}$$

Question 12:

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) red
(ii) black

Answer 12:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{red} \mathrm{balls} = 3 \\ \mathrm{Number} \mathrm{of} \mathrm{black} \mathrm{balls} = 5 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{balls} = 3 + 5 = 8 \\ \left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{drawing} \mathrm{a} \mathrm{red} \mathrm{ball}. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{3}{8} \\ \left(\mathrm{ii}\right) \\ \mathrm{Let} \mathrm{B} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{drawing} \mathrm{a} \mathrm{black} \mathrm{ball}. \\ \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{5}{8} \end{gathered}$$

Question 13:

A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be
(i) red
(ii) white
(iii) not green

Answer 13:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{red} \mathrm{marbles} = 5 \\ \mathrm{Number} \mathrm{of} \mathrm{white} \mathrm{marbles} = 8 \\ \mathrm{Number} \mathrm{of} \mathrm{green} \mathrm{marbles} = 4 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{marbles} \mathrm{in} \mathrm{the} \mathrm{bag} = 5 + 8 + 4 = 17 \\ \therefore \mathrm{Total} \mathrm{number} \mathrm{outcomes} = 17 \\ \left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{drawing} \mathrm{a} \mathrm{red} \mathrm{ball}. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{5}{17} \\ \left(\mathrm{ii}\right) \\ \mathrm{Let} \mathrm{B} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{drawing} \mathrm{a} \mathrm{white} \mathrm{ball}. \\ \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{8}{17} \\ \left(\mathrm{iii}\right) \\ \mathrm{Let} \mathrm{C} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{drawing} \mathrm{a} \mathrm{green} \mathrm{ball}. \\ \therefore \mathrm{P}\left(\mathrm{C}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{4}{17} \\ \mathrm{Now}, \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{not} \mathrm{drawing} \mathrm{a} \mathrm{green} \mathrm{ball} \mathrm{is}: \\ \mathrm{P}\left(\overline{\mathrm{C}}\right) = 1 - \mathrm{P}\left(\mathrm{C}\right) = 1 - \frac{4}{17} = \frac{13}{17} \end{gathered}$$

Page-26.16

Question 14:

If you put 21 consonants and 5 vowels in a bag. What would carry greater probability? Getting a consonant or a vowel? Find each probability.

Answer 14:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{consonants} = 21 \\ \mathrm{Number} \mathrm{of} \mathrm{vowels} = 5 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{possible} \mathrm{outcomes} = 21 + 5 = 26 \\ \mathrm{Let} \mathrm{C} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{consonant} \mathrm{and} \mathrm{V} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{vowel}. \\ \therefore \mathrm{P}\left(\mathrm{C}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{21}{26} \\ \mathrm{And}, \mathrm{P}\left(\mathrm{V}\right)=\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{5}{26} \\ \mathrm{Thus}, \mathrm{the} \mathrm{consonants} \mathrm{have} \mathrm{a} \mathrm{greater} \mathrm{probability}. \end{gathered}$$

Question 15:

If we have 15 boys and 5 girls in a class which carries a higher probability? Getting a copy belonging to a boy or a girl. Can you give it a value?

Answer 15:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{boys} \mathrm{in} \mathrm{the} \mathrm{class} = 15 \\ \mathrm{Number} \mathrm{of} \mathrm{girls} \mathrm{in} \mathrm{the} \mathrm{class} = 5 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students} \mathrm{in} \mathrm{the} \mathrm{class} = 15 + 5 = 20 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{possible} \mathrm{outcomes} = 20 \\ \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{boys} \mathrm{is} \mathrm{more} \mathrm{than} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{girls}, \mathrm{boys} \mathrm{will} \mathrm{have} \mathrm{a} \mathrm{higher} \mathrm{probability}. \\ \mathrm{Hence}, \mathrm{there} \mathrm{is} \mathrm{the} \mathrm{higher} \mathrm{probability} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{copy} \mathrm{belonging} \mathrm{to} \mathrm{a} \mathrm{boy}. \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{boy}\text{'}\mathrm{s} \mathrm{copy} \mathrm{and} \mathrm{B} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{girl}\text{'}\mathrm{s} \mathrm{copy}. \\ \therefore \mathrm{P}\left(\mathrm{A}\right) =\frac{15}{20}=\frac{3}{4} \\ \mathrm{And}, \mathrm{P}\left(\mathrm{B}\right) =\frac{5}{20}=\frac{1}{4} \end{gathered}$$

Question 16:

If you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is
(i) white?
(ii) black?

Answer 16:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{pairs} \mathrm{of} \mathrm{white} \mathrm{socks} = 6 \\ \mathrm{Number} \mathrm{of} \mathrm{pairs} \mathrm{of} \mathrm{black} \mathrm{socks} = 3 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{pairs} \mathrm{of} \mathrm{socks} = 6 + 3 = 9 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{possible} \mathrm{outcomes} = 9 \\ \left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{A} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{pair} \mathrm{of} \mathrm{white} \mathrm{socks}. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{6}{9}=\frac{2}{3} \\ \left(\mathrm{ii}\right) \\ \mathrm{Let} \mathrm{B} \mathrm{be} \mathrm{the} \mathrm{event} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{pair} \mathrm{of} \mathrm{black} \mathrm{socks}. \\ \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{3}{9}=\frac{1}{3} \end{gathered}$$

Question 17:

If you have a spinning wheel with 3-green sectors, 1-blue sector and 1-red sector. What is the probability of getting a green sector? Is it the maximum?

Answer 17:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{green} \mathrm{sectors} \mathrm{in} \mathrm{the} \mathrm{wheel} = 3 \\ \mathrm{Number} \mathrm{of} \mathrm{blue} \mathrm{sectors} \mathrm{in} \mathrm{the} \mathrm{wheel} = 1 \\ \mathrm{Number} \mathrm{of} \mathrm{red} \mathrm{sectors} \mathrm{in} \mathrm{the} \mathrm{wheel} = 1 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{sectors} \mathrm{in} \mathrm{the} \mathrm{wheel} = 3 + 1 + 1 = 5 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{possible} \mathrm{outcomes} = 5 \\ \mathrm{Let} \mathrm{A}, \mathrm{B} \mathrm{and} \mathrm{C} \mathrm{be} \mathrm{the} \mathrm{events} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{green}, \mathrm{blue} \mathrm{and} \mathrm{red} \mathrm{sector}, \mathrm{respectively}. \\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{3}{5}, \mathrm{P}\left(\mathrm{B}\right)=\frac{1}{5} \mathrm{and} \mathrm{P}\left(\mathrm{C}\right)=\frac{1}{5} \\ \mathrm{Hence}, \mathrm{the} \mathrm{probability} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{green} \mathrm{sector} \mathrm{is} \mathrm{the} \mathrm{maximum}. \end{gathered}$$

Question 18:

When two dice are rolled:
(i) List the outcomes for the event that the total is odd.
(ii) Find probability of getting an odd total.
(iii) List the outcomes for the event that total is less than 5.
(iv) Find the probability of getting a total less than 5?

Answer 18:

$$\begin{gathered} \mathrm{Possible} \mathrm{outcomes} \mathrm{when} \mathrm{two} \mathrm{dice} \mathrm{are} \mathrm{rolled}: \\ \mathrm{S}=\left\{\left(1,1\right), \left(1,2\right), \left(1,3\right), \left(1,4\right),\cdots ,\left(6,5\right), \left(6,6\right)\right\} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{possible} \mathrm{outcomes} \mathrm{in} \mathrm{the} \mathrm{sample} \mathrm{space} \mathrm{is} 36. \\ \left(\mathrm{i}\right) \\ \mathrm{The} \mathrm{outcomes} \mathrm{for} \mathrm{the} \mathrm{event} \mathrm{that} \mathrm{the} \mathrm{total} \mathrm{is} \mathrm{odd}: \\ \mathrm{E}=\left\{\left(1,2\right), \left(1,4\right), \left(1,6\right), \left(2,1\right), \left(2,3\right), \left(2,5\right), \left(3,2\right), \left(3,4\right), \left(3,6\right), \left(4,1\right), \left(4,3\right), \left(4,5\right), \left(5,2\right), \left(5,4\right), \left(5,6\right), \left(6,1\right), \left(6,3\right), \left(6,5\right)\right\} \\ \left(\mathrm{ii}\right) \\ \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 18. \\ \therefore \mathrm{P}\left(\mathrm{E}\right)=\frac{18}{36}=\frac{1}{2} \\ \left(\mathrm{iii}\right) \\ \mathrm{The} \mathrm{outcomes} \mathrm{for} \mathrm{the} \mathrm{event} \mathrm{that} \mathrm{total} \mathrm{is} \mathrm{less} \mathrm{than} 5: \\ \mathrm{B}=\left\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(2,1\right),\left(2,2\right),\left(3,1\right)\right\} \\ \left(\mathrm{iv}\right) \\ \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} \mathrm{is} 6. \\ \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{6}{36}=\frac{1}{6} \end{gathered}$$