## RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS

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

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

Question 1.

Define a trial.

Solution:

When we perform an experiment, it is called a trial of the experiment.

Question 2.

Define an elementary event.

Solution:

An outcome of a trial of an experiment is called an elementary event.

Question 3.

Define an event.

Solution:

An event association to a random experiment is said to occur in a trial.

Question 4.

Define probability of an event.

Solution:

In n trials of a random experiment if an event A happens m times, then probability of happening

of A is given by P(A) = \(\frac { m }{ n } \)

Question 5.

A bag contains 4 white balls and some red balls. If the probability of drawing a white ball from the bag is \(\frac { 2 }{ 5 } \), find the number of red balls in the bag

Solution:

No. of white balls = 4

Let number of red balls = x

Then total number of balls (n) = 4 white + x red = (4 + x) balls

Question 6.

A die is thrown 100 times. If the probability of getting an even number is \(\frac { 2 }{ 5 } \). How many times an odd number is obtained?

Solution:

Total number of a die is thrown = 100

Let an even number comes x times, then probability of an even number = \(\frac { x }{ 100 } \)

Question 7.

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes

Find the probability of getting at most two heads.

Solution:

Total number of three coins are tossed (n) = 200

Getting at the most 2 heads (m) = 72 + 77 + 28 = 177

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 177 }{ 200 } \)

Question 8.

In the Q. No. 7, what is the probability of getting at least two heads?

Solution:

Total number of possible events = 200

No. of events getting at the least = 2 heads (m) = 23 + 72 = 95

Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 95 }{ 200 } \) = \(\frac { 19 }{ 40 } \)

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

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