$\begin{array}{1 1}44/91 \\ 12/91 \\ 11/91 \\ 15/91 \end{array} $

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- If A and B are independant events, \(P(A\cap\;B)=P(A)\;P(B)\)

Given a box of 15 oranges, let A be the event that 12 oranges picked are good and 3 are bad. Then P(A) = $\large \frac{12}{15}$.

Now that the first orange is picked, we have 14 left. Let B be the event that the 11 oranges picked are good. Then P(B) = $\large \frac{11}{14}$

Let C be the event that out of the remaining 13 oranges, 10 are good. Then P(C) = $\large \frac{10}{13}$

Since, A, B and C are independant events, we can apply the rule that \(P(A\cap\;B)=P(A)\;P(B)\)

Therefore, by extension P (A $\cap$ B $\cap$ C ) = P(A) $\times$ P(B) $\times$ P(C).

$\Rightarrow$ P (A $\cap$ B $\cap$ C ) = $\large \frac{12}{15}$ $\times$ $\large \frac{11}{14}$ $\times$ $\large \frac{10}{13}$ = $\large\frac{44}{91}$

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