Question

A basket contains $20$ apples and $10$ oranges out of which $5$ apples and $3$ oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer 1

Total:- $10$ Oranges $20$ Apples

Rotten:- $3$ $5$

Good fruits:- $7$ $15$

Let $A=$ Event of selecting one apple

& $B=$ Event of selecting one good fruit

We have to find $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\bigcap B)$ ......(1)

Now, $P\left(A\right)=\frac{{}^{20}{C}_1}{{}^{30}{C}_1}=\frac{2}{3}$ {as there are $20$ apples}

$P\left(B\right)=\frac{{}^{22}{C}_1}{{}^{30}{C}_1}=\frac{11}{15}$ {as there are total $7+15=22$ good fruits}

$P(A\bigcap B)=\frac{15C_1}{{}^{30}{C}_1}=\frac{1}{2}$ {as there are $15$ non-defective apples}

Substituting the values in equation (1),

$P(A\cup B)=\frac{2}{3}+\frac{11}{15}-\frac{1}{2}$

The probability that the fruit is either an apple or a good fruit is $\frac{9}{10}$.

Note: “Here we are given Either Or which means Exclusive OR i.e. We have to find probability of getting one apple or one good fruit but not apple and good fruit, both simultaneously”.