__Frequently Asked Questions on Chapter 15- Probability__

__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