RS AGGARWAL CLASS 9 CHAPTER 19 PROPABILITY MCQ

 MULTIPLE CHOICE QUESTIONS 

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Question 3:

80 bulbs are selected at random from a lot and their lifetime is hours is recorded as under.

Lifetime (in hours)	300	500	700	900	1100
Frequency	10	12	23	25	10

One bulb is selected at random from the lot. What is the probability that its life is 1150?
(a) $\frac{1}{80}$$\frac{1}{80}$

(b) $\frac{7}{16}$$\frac{7}{16}$

(c) 1

(d) 0

Answer 3:

(d) 0
Maximum lifetime a bulb has is 1100 hours. There is no bulb with lifetime 1150 hours.

Question 4:

In a survey of 364 children aged 19 – 36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he / she does not like to eat potato chips is
(a) $\frac{1}{4}$$\frac{1}{4}$

(b) $\frac{1}{2}$$\frac{1}{2}$

(c) $\frac{3}{4}$$\frac{3}{4}$

(d) $\frac{4}{5}$$\frac{4}{5}$

Answer 4:

Total number of children surveyed = 364

Number of children who liked to eat potato chips = 91

Number of children who do not liked to eat potato chips = 364 − 91 = 273

∴ P(Child does not like to eat potato chips) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{children}\mathrm{who}\mathrm{do}\mathrm{not}\mathrm{liked}\mathrm{to}\mathrm{eat}\mathrm{potato}\mathrm{chips}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{children}\mathrm{surveyed}}=\frac{273}{364}=\frac{3}{4}$$\frac{\mathrm{Number} \mathrm{of} \mathrm{children} \mathrm{who} \mathrm{do} \mathrm{not} \mathrm{liked} \mathrm{to} \mathrm{eat} \mathrm{potato} \mathrm{chips}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{children} \mathrm{surveyed}}=\frac{273}{364}=\frac{3}{4}$

Hence, the correct answer is option (c).

Question 5:

Two coins are tossed 1000 times and the outcomes are recorded as given below:

Number of heads	2	1	0
Frequency	200	550	250

Now, if two coins are tossed at random, what is the probability of getting at most one head?

(a) $\frac{3}{4}$$\frac{3}{4}$

(b) $\frac{4}{5}$$\frac{4}{5}$

(c) $\frac{1}{4}$$\frac{1}{4}$

(d) $\frac{1}{5}$$\frac{1}{5}$

Answer 5:

Total number of times two coins are tossed = 1000

Number of times of getting at most one head = Number of times of getting 0 heads + Number of times of getting 1 head = 250 + 550 = 800

∴ P(Getting at most one head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}\mathrm{at}\mathrm{most}\mathrm{one}\mathrm{head}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{two}\mathrm{coins}\mathrm{are}\mathrm{tossed}}=\frac{800}{1000}=\frac{4}{5}$$\frac{\mathrm{Number} \mathrm{of} \mathrm{times} \mathrm{of} \mathrm{getting} \mathrm{at} \mathrm{most} \mathrm{one} \mathrm{head}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{times} \mathrm{two} \mathrm{coins} \mathrm{are} \mathrm{tossed}}=\frac{800}{1000}=\frac{4}{5}$

Hence, the correct answer is option (b).

Question 6:

80 bulbs are selected at random from a lot and their lifetime in hours is recorded as under.

Lifetime (in hours)	300	500	700	900	1100
Frequency	10	12	23	25	10

One bulb is selected at random from the lot. What is the probability that the selected bulb has a life more than 500 hours?

(a) $\frac{27}{40}$$\frac{27}{40}$

(b) $\frac{29}{40}$$\frac{29}{40}$

(c) $\frac{5}{16}$$\frac{5}{16}$

(d) $\frac{11}{40}$$\frac{11}{40}$

Answer 6:

(b) $\frac{29}{40}$$\frac{29}{40}$

Explanation:
 Total number of bulbs in the lot  = 80
 Number of bulbs with life time of more than 500 hours = (23 + 25 + 10) = 58

 Let E be the event that the chosen bulb's life time is more than 500 hours.
 

∴ Required probability = P(E) =  $\frac{58}{80}=\frac{29}{40}$$\frac{58}{80} = \frac{29}{40}$

Question 7:

To know the opinion of the students about the subject Sanskrit, a survey of 200 students was conducted. The data is recorded as under.

Opinion	like	dislike
Number of students	135	65

What is the probability that a student chosen at random does not like it?

(a) $\frac{13}{27}$$\frac{13}{27}$

(b) $\frac{27}{40}$$\frac{27}{40}$

(c) $\frac{13}{40}$$\frac{13}{40}$

(d) $\frac{27}{13}$$\frac{27}{13}$

Answer 7:

Total number of students surveyed = 200

Number of students who does not like the subject Sanskrit = 65

∴ P(Student chosen at random does not like the subject Sanskrit) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{does}\mathrm{not}\mathrm{like}\mathrm{the}\mathrm{subject}\mathrm{Sanskrit}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}\mathrm{surveyed}}=\frac{65}{200}=\frac{13}{40}$$\frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{who} \mathrm{does} \mathrm{not} \mathrm{like} \mathrm{the} \mathrm{subject} \mathrm{Sanskrit}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students} \mathrm{surveyed}}=\frac{65}{200}=\frac{13}{40}$

Hence, the correct answer is option (c).

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Question 8:

A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting a head?
(a) $\frac{7}{12}$$\frac{7}{12}$

(b) $\frac{12}{7}$$\frac{12}{7}$

(c) $\frac{5}{12}$$\frac{5}{12}$

(d) $\frac{1}{25}$$\frac{1}{25}$

Answer 8:

(c) $\frac{5}{12}$$\frac{5}{12}$
​
Explanation:
Total number of trials = 60
Number of times tail appears = 35

Number of times head appears = 60 − 35 = 25

Let E be the event of getting a head.
  ∴ P(getting a head) = P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{25}{60}=\frac{5}{12}$$\frac{\mathrm{Number} \mathrm{of} \mathrm{times} \mathrm{head} \mathrm{appears}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{25}{60} = \frac{5}{12}$

Question 9:

It is given that the probability of winning a game is 0.7. What is the probability of losing the game?
(a) 0.8
(b) 0.3
(c) 0.35
(d) 0.15

Answer 9:

(b) 0.3

Explanation:
Let E be the event of winning the game. Then,
P(E) =  0.7
P(not E) = P(losing the game) = 1− ​P(E)  ⇒ 1− 0.7 = 0.3

Question 10:

In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. What is the probability that in a given delivery, the ball does not hit the boundary?
(a) $\frac{1}{4}$$\frac{1}{4}$

(b) $\frac{1}{5}$$\frac{1}{5}$

(c) $\frac{4}{5}$$\frac{4}{5}$

(d) $\frac{3}{4}$$\frac{3}{4}$

Answer 10:

(c) $\frac{4}{5}$$\frac{4}{5}$

Explanation: 
Total number of balls faced = 30
Number of times the ball hits the boundary = 6
Number of times the ball does not hit the boundary = (30 − 6 )= 24

Let E be the event that the ball does not hit the boundary. Then,

 P(E) =  $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{ball}\mathrm{does}\mathrm{not}\mathrm{hit}\mathrm{the}\mathrm{boundary}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{balls}}=\frac{24}{30}=\frac{4}{5}$$\frac{\mathrm{Number} \mathrm{of} \mathrm{times} \mathrm{ball} \mathrm{does} \mathrm{not} \mathrm{hit} \mathrm{the} \mathrm{boundary}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{balls}} = \frac{24}{30} = \frac{4}{5}$

Question 11:

A bag contains 16 cards bearing numbers 1, 2, 3, ..., 16 respectively. One card is chosen at random. What is the probability that the chosen card bears a number divisible by 3?
(a) $\frac{3}{16}$$\frac{3}{16}$

(b) $\frac{5}{16}$$\frac{5}{16}$

(c) $\frac{11}{16}$$\frac{11}{16}$

(d) $\frac{13}{16}$$\frac{13}{16}$

Answer 11:

(b) $\frac{5}{16}$$\frac{5}{16}$

Explanation:
Total number of cards in the bag =  16
Numbers on the cards that are divisible by 3 are 3, 6, 9, 12 and 15.

Number of cards with numbers divisible by 3 = 5

​Let E be the event that the chosen card bears a number divisible by 3.

 

∴ Required probability = P(E) = $\frac{5}{16}$$\frac{5}{16}$

Question 12:

A bag contains 5 red, 8 black and 7 white balls. One ball is chosen at random. What is the probability that the chosen ball is black?
(a) $\frac{2}{3}$$\frac{2}{3}$

(b) $\frac{2}{5}$$\frac{2}{5}$

(c) $\frac{3}{5}$$\frac{3}{5}$

(d) $\frac{1}{3}$$\frac{1}{3}$

Answer 12:

(b) $\frac{2}{5}$$\frac{2}{5}$

Explanation:
Total number of balls in the bag =  5 + 8 + 7 = 20
 Number of black balls  = 8

 Let E be the event that the chosen ball is black.
 

∴ Required probability = P(E) = $\frac{8}{20}=\frac{2}{5}$$\frac{8}{20} = \frac{2}{5}$

Question 13:

The outcomes of 65 throws of a dice were noted as shown below:

Outcome	1	2	3	4	5	6
Number of times	8	10	12	16	9	10

A dice is thrown at random. What is the probability of getting a prime number?
(a) $\frac{3}{35}$$\frac{3}{35}$

(b) $\frac{3}{5}$$\frac{3}{5}$

(c) $\frac{31}{65}$$\frac{31}{65}$

(d) $\frac{36}{65}$$\frac{36}{65}$

Answer 13:

(c) $\frac{31}{65}$$\frac{31}{65}$

Explanation:
Total number of throws = 65
Let E be the event of getting a prime number. 
Then, E contains 2, 3 and 5, i.e. three numbers.

∴ P(getting a prime number) = P(E)  = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{prime}\mathrm{number}\text{s}\mathrm{occur}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{throws}}=\frac{(10+12+9)}{65}=\frac{31}{65}$$\frac{\mathrm{Number} \mathrm{of} \mathrm{times} \mathrm{prime} \mathrm{number}\text{s} \mathrm{occur}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{throws}} = \frac{(10+12+9)}{65} = \frac{31}{65}$

Question 14:

In 50 throws of a dice, the outcomes were noted as shown below:

Outcome	1	2	3	4	5	6
Number of times	8	9	6	7	12	8

A dice is thrown at random. What is the probability of getting an even number?
(a) $\frac{12}{25}$$\frac{12}{25}$

(b) $\frac{3}{50}$$\frac{3}{50}$

(c) $\frac{1}{8}$$\frac{1}{8}$

(d) $\frac{1}{2}$$\frac{1}{2}$

Answer 14:

(a) $\frac{12}{25}$$\frac{12}{25}$

Explanation:
Total number of trials =  50
Let E be the event of getting an even number.
Then, E contains 2, 4 and 6, i.e. 3 even numbers.

∴ P(getting an even number) = P(E)  = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{even}\mathrm{number}s\mathrm{appear}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{throws}}=\frac{(9+7+8)}{50}=\frac{24}{50}=\frac{12}{25}$$\frac{\mathrm{Number} \mathrm{of} \mathrm{times} \mathrm{even} \mathrm{number}s \mathrm{appear}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{throws}} = \frac{(9+7+8)}{50} = \frac{24}{50} =\frac{12}{25}$

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Question 15:

The table given below shows the month of birth of 36 students of a class.

Month of birth	Jan	Feb	Mar	Apr	May	June	July	Aug	Sept	Oct	Nov	Dec
No. of students	4	3	5	0	1	6	1	3	4	3	4	2

A student is chosen at random from the class. What is the probability that the chosen student was born in October?
(a) $\frac{1}{3}$$\frac{1}{3}$

(b) $\frac{2}{3}$$\frac{2}{3}$

(c) $\frac{1}{4}$$\frac{1}{4}$

(d) $\frac{1}{12}$$\frac{1}{12}$

Answer 15:

(d) $\frac{1}{12}$$\frac{1}{12}$

Explanation: 
Total number of students = 36
Number of students born in October = 3

Let E be the event that the chosen student was born in October. Then,

 P(E) =  $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{born}\mathrm{in}\mathrm{October}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}}=\frac{3}{36}=\frac{1}{12}$$\frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{born} \mathrm{in} \mathrm{October}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}} = \frac{3}{36} = \frac{1}{12}$

Question 16:

Two coins are tossed simultaneously 600 times to get 2 heads : 234 times, 1 head : 206 times, 0 head : 160 times.
If two coins are tossed at random, what is the probability of getting at least one head?

(a) $\frac{103}{300}$$\frac{103}{300}$     

(b) $\frac{39}{100}$$\frac{39}{100}$

(c) $\frac{11}{15}$$\frac{11}{15}$

(d) $\frac{4}{15}$$\frac{4}{15}$

Answer 16:

Number of times two coins are tossed simultaneously = 600

Number of times of getting at least one head = Number of times of getting 1 head + Number of times of getting 2 heads = 206 + 234 = 440

∴ P(Getting at least one head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{head}}{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{two}\mathrm{coins}\mathrm{are}\mathrm{tossed}\mathrm{simultaneously}}=\frac{440}{600}=\frac{11}{15}$$\frac{\mathrm{Number} \mathrm{of} \mathrm{times} \mathrm{of} \mathrm{getting} \mathrm{at} \mathrm{least} \mathrm{one} \mathrm{head}}{\mathrm{Number} \mathrm{of} \mathrm{times} \mathrm{two} \mathrm{coins} \mathrm{are} \mathrm{tossed} \mathrm{simultaneously}}=\frac{440}{600}=\frac{11}{15}$

Hence, the correct answer is option (c).