Three mangoes and three apples are in a box. | vedclass

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(B) Total number of fruits in the box $= 3 	ext{ (mangoes)} + 3 	ext{ (apples)} = 6 	ext{ fruits}.$
We need to choose $2$ fruits out of $6$. The t

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$\frac{3}{5}$

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

Three mangoes and three apples are in a box. If two fruits are chosen at random,the probability that one is a mango and the other is an apple is

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