- Home
- Standard 11
- Mathematics
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing $15$ oranges out of which $12$ are good and $3$ are bad ones will be approved for sale.
$\frac{44}{91}$
$\frac{44}{91}$
$\frac{44}{91}$
$\frac{44}{91}$
Solution
Let $A, B,$ and $C$ be the respective events that the first, second, and the third drawn orange is good.
Therefore, probability that first drawn orange is good, $\mathrm{P}(\mathrm{A})=\frac{12}{15}$
The oranges are not replaced.
Therefore, probability of getting second orange good, $\mathrm{P}(\mathrm{B})=\frac{11}{14}$
Similarly, probability of getting third orange good, $\mathrm{P}(\mathrm{C})=\frac{10}{13}$
The box is approved for sale, if all the three oranges are good.
Thus, probability of getting all the oranges good $=\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}=\frac{44}{91}$
Therefore, the probability that the box is approved for sale is $\frac{44}{91}$.
