Solution:
Given: A box of oranges.
Let A, B, and C represent the events that occur when the first, second, and third drawn oranges are all excellent.
Now, $P(A)=P$ (good orange in first draw) $=12 / 15$.
Because the second orange is not replaced, the total number of excellent oranges will now be 11 and the total number of oranges will be 14, which is the conditional probability of B if A has already occurred.
Now, $P(B / A)=P($ good orange in second draw $)=11 / 14$
Because the third orange is not replaced, the total number of excellent oranges is now 10 and the total number of orangs is 13, which is the conditional probability of C provided that $A$ and $B$ have already occurred.
Now, $\mathrm{P}(\mathrm{C} / \mathrm{AB})=\mathrm{P}$ (good orange in third draw) $=10 / 13$
As a result, there’s a strong chance that all of the oranges are excellent.
$\Rightarrow P(A \cap B \cap C)=12 / 15 \times 11 / 14 \times 10 / 13=44 / 91$
As a result, the likelihood of a box being authorised for sale is high. $=44 / 91$