Bag A contains 6 Green and 8 Red balls and B contains 9 Green and 5 Red balls. A card is drawn at random from a well shuffled pack of 52 playing cards. If it is a spade, two balls are drawn at random from bag A, otherwise two balls are drawn at random from bag B. If the two balls drawn are found to be the same colour, then the probability that they are drawn from bag A is

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$\frac{43}{181}$

$\frac{43}{138}$

$\frac{48}{131}$

$\frac{1}{4}$

answer is A.

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Detailed Solution

	A	B
Green	6	9
Red	8	5

$E_{1} =$ a card drawn from pack of chords its spade then bag A selected.

$E_{2} =$ bag B selected a card is draw its not spade.

$P (E_{1}) = \frac{13}{52} = \frac{1}{4} P (E_{2}) = \frac{3}{4}$

A=2 balls drawn both are same colour

$\begin{array}{rcl} P (\frac{A}{E_{1}}) & = & \frac{{}^{6}{C_{2} + {}^{8}{C_{2}}}}{{}^{14}C_{2}} P (\frac{A}{E_{2}}) = \frac{{}^{9}{C_{2} + {}^{5}{C_{2}}}}{{}^{14}C_{2}} \end{array}$

$\begin{array}{rcl} P (\frac{E_{1}}{A}) & = \end{array}$ $\begin{array}{rcl} \frac{\frac{1}{4} \times \frac{{}^{6}{C_{2} + {}^{8}{C_{2}}}}{{}^{14}C_{2}}}{\frac{1}{4} \times \frac{{}^{6}{C_{2} + {}^{8}{C_{2}}}}{{}^{14}C_{2}} + \frac{3}{4} \times (\frac{{}^{9}{C_{2} + {}^{5}{C_{2}}}}{{}^{14}C_{2}})} \end{array}$