Question

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red, (ii) not red.

Answer 1

Hint: The probability of an event X is given by the ratio of the number of favourable outcomes corresponding to that event and the total number of possible outcomes. Using this formula, we can solve this question.

Complete step-by-step answer:
Before proceeding with the question, we must know the formula that will be required to solve this question.

Let us assume an event X. From probability, the probability of this event X is given by dividing the number of favourable outcomes corresponding to the event X by the total number of outcomes that are possible i.e. the number of elements in the sample space. Mathematically,
$P\left( X \right)=\dfrac{n\left( X \right)}{n\left( S \right)}$ . . . . . . . . . . . . . (1)

(i) It is given that a bag contains 3 red balls and 5 black balls and we are required to calculate the probability that the ball drawn is red.

Number of elements in the sample space i.e. n(S) = 3 + 5 = 8.

Let us assume the X is an event which denotes that the ball drawn is red. Since there are 3 red balls in total, we can say that n(X) = 3.

Using formula (1), we get,
$P\left( X \right)=\dfrac{3}{8}$

(ii) It is given that a bag contains 3 red balls and 5 black balls and we are required to calculate the probability that the ball drawn is not red. This means that the ball drawn should be black in colour.

Number of elements in the sample space i.e. n(S) = 3 + 5 = 8.

Let us assume the X is an event which denotes that the ball drawn is black. Since there are 5 black balls in total, we can say that n(X) = 5.

Using formula (1), we get,
$P\left( X \right)=\dfrac{5}{8}$
Hence, the probability in part (i) is $\dfrac{3}{8}$ and the probability in part (ii) is $\dfrac{5}{8}$.


Note: The probability in part (ii) can be also found by subtracting the probability obtained in part (i) from 1. This is because in probability, we have a formula P(X) + P(X’) = 1 where X’ is the complement of X. In this question, the event in part (ii) is the complement of the event in part (i). That is why, we can use this formula to find the probability in part (ii).