Bag \(A\) contains 6 Green and 8 Red balls and bag \(B\) contains 9 Green and 5 Red balls. A card is drawn…
Bag \(A\) contains 6 Green and 8 Red balls and bag \(B\) 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 of the same colour, then the probability that they are drawn from bag \(A\) is
\(\frac{43}{181}\)
\(\frac{1}{4}\)
\(\frac{48}{131}\)
\(\frac{43}{138}\)
Solution
According to given informations, the required probability
\(\begin{aligned}
= & \frac{\frac{1}{4}\left(\frac{{ }^6 C_2+{ }^8 C_2}{{ }^{14} C_2}\right)}{\frac{1}{4}\left(\frac{{ }^6 C_2+{ }^8 C_2}{{ }^{14} C_2}\right)+\frac{3}{4}\left(\frac{{ }^9 C_2+{ }^5 C_2}{{ }^{14} C_2}\right)} \\
= & \frac{(6 \times 5)+(8 \times 7)}{[(6 \times 5)+(8 \times 7)]+3[(9 \times 8)+(5 \times 4)]} \\
= & \frac{30+56}{(30+56)+3(72+20)}=\frac{86}{86+276}=\frac{86}{362}=\frac{43}{181}
\end{aligned}\)
Hence, option (1) is correct.