Question

Shown below are two baskets with grey and black balls.

Abhishek is playing a game with his friend where he has to close his eyes and pick a black ball from one of the baskets in one trial.

He said "I will try with basket $2$ as it has a higher number of black balls than basket $1$ and hence the probability of picking a black ball from basket $2$ is higher."

Is Abhishek's statement correct? Justify your answer.

Accepted Answer

No, Abhishek's statement is not correct.

Total balls in basket $1$ $= 3 Grey balls + 4 Black balls$

Number of possible outcomes in basket $1$ $= 3 + 4 = 7$

Number of black balls in basket $1$ $= 4$

Hence, the number of favourable outcomes $= 4$

We have, $P (E) = \frac{Number of favourable outcomes}{Number of possible outcomes}$

$\Rightarrow P (black ball from basket 1) = \frac{4}{7}$ $. . . (1)$

Now, total balls in basket $2$ $= 6 Grey balls + 8 Black balls$

Number of possible outcomes in basket $2$ $= 6 + 8 = 14$

Number of black balls in basket $2$ $= 8$

Hence, the number of favourable outcomes $= 8$

We have, $P (E) = \frac{Number of favourable outcomes}{Number of possible outcomes}$

$\Rightarrow P (black ball from basket 2) = \frac{8}{14} = \frac{2}{7}$ $. . . (2)$

From equations $(1) and (2)$ ,

The probability of picking a black ball from basket $1$ is same as basket $2$ .

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