A basket contains 3 mangoes and 3 apples. If | vedclass

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(B) Total number of fruits = $3$ (mangoes) + $3$ (apples) = $6$ fruits. We need to select $2$ fruits out of $6$. The total number of ways to select $2

Vedclass · Accepted Answer

$3/5$

Vedclass · Answer

(B) Total number of fruits = $3$ (mangoes) + $3$ (apples) = $6$ fruits. We need to select $2$ fruits out of $6$. The total number of ways to select $2$ fruits is given by $^6C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$. We want to select $1$ mango out of $3$ and $1$ apple out of $3$. The number of favorable ways is $^3C_1 	imes ^3C_1 = 3 	imes 3 = 9$. The probability $P$ is given by $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{9}{15} = \frac{3}{5}$.

$A$ basket contains $3$ mangoes and $3$ apples. If two fruits are selected at random,what is the probability of getting one mango and one apple?

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