Course Content
Class 10th Maths
0/86
Online Class For 10th Standard Students (CBSE) (English Medium)
About Lesson

Frequently Asked Questions on Chapter 15- Probability

Complete the following statements Probability of an event E + Probability of the event ‘not E’ = ___________ ?

Probability of an event E + Probability of the event ‘not E’ = 1.

Which of the following experiments have equally likely outcomes Explain A driver attempts to start a car. The car starts or does not start?

This statement does not have equally likely outcomes as the car may or may not start depending upon various factors like fuel, etc.

Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.

A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out an orange flavoured candy?

We know that the bag only contains lemon-flavoured candies. So, The no. of orange flavoured candies = 0 ∴ The probability of taking out orange flavoured candies = 0/1 = 0

It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Let the event wherein 2 students having the same birthday be E Given, P(E) = 0.992 We know, P(E) + P(not E) = 1 Or, P(not E) = 1 – 0.992 = 0.008 ∴ The probability that the 2 students have the same birthday is 0.008

Tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game? why?

Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.

What is the probability of ‘not E’, if P(E) = 0.05?

We know that,

P(E) + P(not E) = 1

It is given that, P(E) = 0.05

So, P(not E) = 1 – P(E)

Or, P(not E) = 1 – 0.05

Hence, P(not E) = 0.95

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?

The Total no. of marbles = 5 + 8 + 4 = 17

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of red marbles= 5

P (red marbles) = 5/17 = 0.29

(ii) Total number of white marbles= 8

P (white marbles) = 8/17 = 0.47

(iii) Total number of green marbles = 4

P (green marbles) = 4/17 = 0.23

Hence, P (not green) = 1 – P (green marbles)

= 1 – (4/17)

= 0.77

Exercise Files
No Attachment Found
No Attachment Found
Wisdom Academy