Bag A contains 6 Green and 8 Red balls and ba | vedclass

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(A) Let $E_1$ be the event that a spade is drawn,and $E_2$ be the event that a spade is not drawn. Then $P(E_1) = \frac{13}{52} = \frac{1}{4}$ and $P(

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$\frac{43}{181}$

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(A) Let $E_1$ be the event that a spade is drawn,and $E_2$ be the event that a spade is not drawn. Then $P(E_1) = \frac{13}{52} = \frac{1}{4}$ and $P(E_2) = 1 - \frac{1}{4} = \frac{3}{4}$.Let $S$ be the event that two balls of the same colour are drawn.For bag $A$ ($6$ Green,$8$ Red,total $14$): $P(S|E_1) = \frac{{}^6C_2 + {}^8C_2}{{}^{14}C_2} = \frac{15 + 28}{91} = \frac{43}{91}$.For bag $B$ ($9$ Green,$5$ Red,total $14$): $P(S|E_2) = \frac{{}^9C_2 + {}^5C_2}{{}^{14}C_2} = \frac{36 + 10}{91} = \frac{46}{91}$.Using Bayes' Theorem,the probability that the balls were drawn from bag $A$ given they are the same colour is:$P(E_1|S) = \frac{P(E_1)P(S|E_1)}{P(E_1)P(S|E_1) + P(E_2)P(S|E_2)}$$P(E_1|S) = \frac{\frac{1}{4} 	imes \frac{43}{91}}{\frac{1}{4} 	imes \frac{43}{91} + \frac{3}{4} 	imes \frac{46}{91}} = \frac{43}{43 + 3 	imes 46} = \frac{43}{43 + 138} = \frac{43}{181}$.

Box	Red	White	Black
$A$	$1$	$6$	$3$
$B$	$6$	$2$	$2$
$C$	$8$	$1$	$1$
$D$	$0$	$6$	$4$

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Events $E_{1}$ and $E_{2}$ form a partition of the sample space $S$. $A$ is any event such that $P(E_{1}) = P(E_{2}) = \frac{1}{2}$,$P(E_{2} | A) = \frac{1}{2}$,and $P(A | E_{2}) = \frac{2}{3}$. Then $P(E_{1} | A)$ is:

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