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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.

$\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 ) = $\large \frac{12}{15}$ $\times$ $\large \frac{11}{14}$ $\times$ $\large \frac{10}{13}$ = $\large\frac{44}{91}$
answered Jun 19, 2013 by balaji.thirumalai

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