A basket contains 5 apples and 7 oranges and | vedclass

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(C) Let $B_1$ be the first basket and $B_2$ be the second basket. In $B_1$,total fruits = $5 + 7 = 12$. Probability of picking an apple from $B_1$,$P(

Vedclass · Accepted Answer

$\frac{17}{36}$

Vedclass · Answer

(C) Let $B_1$ be the first basket and $B_2$ be the second basket. In $B_1$,total fruits = $5 + 7 = 12$. Probability of picking an apple from $B_1$,$P(A_1) = \frac{5}{12}$. Probability of picking an orange from $B_1$,$P(O_1) = \frac{7}{12}$. In $B_2$,total fruits = $4 + 8 = 12$. Probability of picking an apple from $B_2$,$P(A_2) = \frac{4}{12} = \frac{1}{3}$. Probability of picking an orange from $B_2$,$P(O_2) = \frac{8}{12} = \frac{2}{3}$. We want one apple and one orange. This can happen in two mutually exclusive ways: $1$. Apple from $B_1$ and Orange from $B_2$: $P(A_1) 	imes P(O_2) = \frac{5}{12} 	imes \frac{2}{3} = \frac{10}{36}$. $2$. Orange from $B_1$ and Apple from $B_2$: $P(O_1) 	imes P(A_2) = \frac{7}{12} 	imes \frac{1}{3} = \frac{7}{36}$. Total probability = $\frac{10}{36} + \frac{7}{36} = \frac{17}{36}$.

$A$ basket contains $5$ apples and $7$ oranges and another basket contains $4$ apples and $8$ oranges. If one fruit is picked out at random from each basket,then the probability of getting one apple and one orange is

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