Urn I contains 6 red balls and 4 black balls and Urn II contains 4 red balls and 6 black balls. One ball is drawn at random from Urn I and placed in Urn II. If one ball is drawn at random from Urn II, then the probability that it is a red ball is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{23}{55}\)
Let \(E_1\): the ball moved from Urn I is red, \(E_2\): it is black, and R: the ball drawn from Urn II is red. After the transfer Urn II has 11 balls. By the theorem of total probability,
\(P(R) = P(E_1)P(R \mid E_1)+P(E_2)P(R \mid E_2)\)
If a red ball is moved, Urn II has 5 red out of 11; if a black ball is moved, 4 red out of 11:
\( \displaystyle P(R) = \frac{6}{10} \times \frac{5}{11}+\frac{4}{10} \times \frac{4}{11} \)
\( \displaystyle = \frac{30+16}{110} = \frac{46}{110} \)
\( \displaystyle = \frac{23}{55} \)