Bag I contains 3 black and 2 white balls, Bag II contains 2 black and 4 white balls. A bag is selected at random and then a ball is drawn from it. The probability that the ball drawn is black is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{7}{15}\)
Let \(E_1\) and \(E_2\) be the events that Bag I and Bag II are selected, and B the event that the ball drawn is black.
\(\displaystyle P(E_1) = P(E_2) = \frac{1}{2}\)
\(\displaystyle P(B \mid E_1) = \frac{3}{5},\quad P(B \mid E_2) = \frac{2}{6} = \frac{1}{3}\)
By the theorem of total probability:
\(P(B) = P(E_1)P(B \mid E_1) + P(E_2)P(B \mid E_2)\)
\(\displaystyle = \frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{1}{3}\)
\(\displaystyle = \frac{3}{10} + \frac{1}{6}\)
\(\displaystyle = \frac{9 + 5}{30} = \frac{7}{15}\)
Tip: Adding \(\frac{3}{5} + \frac{1}{3} = \frac{14}{15}\) without the factor \(\frac{1}{2}\) for choosing the bag gives Option (3), a common slip. A probability made of two such cases must be weighted by the chance of each case.
NCERT Exemplar:
Probability
