## RD Sharma Class 9 Solutions Chapter 25 Probability MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 25 Probability MCQS

Other Exercises

- RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1
- RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS
- RD Sharma Class 9 Solutions Chapter 25 Probability MCQS

Mark the correct alternative in each of the following:

Question 1.

The probability of an impossible event is

(a) 1

(b) 0

(c) less than 0

(d) greater than 1

Solution:

The probability of an impossible event is 0 (b)

Question 2.

The probability on a certain event is

(a) 0

(b) 1

(c) greater than 1

(d) less than 1

Solution:

The probability of a certain event is 1 (b)

Question 3.

The probability of an event of a trial is

(a) 1

(b) 0

(c) less than 1

(d) more than 1

Solution:

The probability of an even of a trial is less than 1 (c)

Question 4.

Which of the following cannot be the probability of an event?

(a) \(\frac { 1 }{ 3 } \)

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

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

(d) 1

Solution:

The probability of an event is less than 1

\(\frac { 5 }{ 3 } \) i.e .\(1\frac { 2 }{ 3 } \) is not the probability

Question 5.

Two coins are tossed simultaneously. The probability of getting atmost one head is

(a) \(\frac { 1 }{ 4 } \)

(b) \(\frac { 3 }{ 4 } \)

(c) \(\frac { 1 }{ 2 } \)

(d) \(\frac { 1 }{ 4 } \)

Solution:

Total number of possible events (n) = 2 + 2 = 4

Number of events coming at the most 1 head (m) 2 times + 1 times = 3

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 3 }{ 4 } \) (b)

Question 6.

A coin is tossed 1000 times, if the probability of getting a tail is 3/8, how many times head is obtained?

(a) 525

(b) 375

(c) 625

(d) 725

Solution:

No. of times a coin is tossed (n) = 1000

Probability of getting a tail = \(\frac { 3 }{ 8 } \)

Let No. of tail come = x

Probability P(A) = \(\frac { m }{ n } =\frac { x }{ 1000 } \)

\(\frac { x }{ 1000 } \) = \(\frac { 3 }{ 8 } \)

=> \(\frac { x }{ 1000 } =\frac { 3 }{ 8 } \) => \(\frac { 3X1000 }{ 8 } =3X125\)

=> x = 375

∴ No. of heads are obtained = 1000 – 375 = 625 (c)

Question 7.

A dice is rolled 600 times and the occurrence of the outcomes 1, 2, 3, 4, 5 and 6 are given below:

The probability of getting a prime number is

(a)\(\frac { 1 }{ 3 } \)

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

(c)\(\frac { 49 }{ 60 } \)

(d)\(\frac { 39 }{ 125 } \)

Solution:

Total number of times a dice is rolled (n) = 600

Now total number of times getting a prime number i.e. 2, 3, 5, (m) = 30 + 120 + 50 = 200

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 200 }{ 600 } \) = \(\frac { 1 }{ 3 } \) (a)

Question 8.

The percentage of attendance of different classes in a year in a school is given below:

What is the probability that the class attendance is more than 75%?

(a) \(\frac { 1 }{ 6 } \)

(b) \(\frac { 1 }{ 3 } \)

(c) \(\frac { 5 }{ 6 } \)

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

Solution:

Percentage of attendance of different classes

Total attendance more than 75% (m) VIII,VII and VI = 3 classes

and total number of classes (n) = 6

Probability P(A) = \(\frac { 3 }{ 6 } \) = \(\frac { 1 }{ 2 } \).

Question 9.

A bag contains 50 coins and each coin is marked from 51 to 100. One coin is picked at random.

The probability that the number on the coin is not a prime number, is

(a) \(\frac { 1 }{ 5 } \)

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

(c) \(\frac { 2 }{ 5 } \)

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

Solution:

Total number of coins (n) = 50

Prime numbers between 51 to 100 are 53, 59, 6, 67, 71, 73, 79, 83, 89, 97 = 10

Numbers which are not primes (m) = 50 – 10 = 40

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 40 }{ 50 } \) = \(\frac { 4 }{ 5 } \)(d)

Question 10.

In a football match, Ronaldo makes 4 pals from 10 penalty kids. The probability of converting a penalty kick into a goal by Ronaldo,is

(a) \(\frac { 1 }{ 4 } \)

(b) \(\frac { 1 }{ 6 } \)

(c) \(\frac { 1 }{ 3 } \)

(d) \(\frac { 2 }{ 5 } \)

Solution:

No. of penalty kicks (n) = 10

No. of goal scored (m) = 4

Probability of converting a penally Into goals P(A) = \(\frac { 4 }{ 10 } \) = \(\frac { 2 }{ 5 } \)(d)

Hope given RD Sharma Class 9 Solutions Chapter 25 Probability MCQS are helpful to complete your math homework.

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