Question

A bag A contains 3 red, 2 white and 2 black balls and another bag B contains 1 red, 2 white and 4 black balls. A die is thrown to select a bag. If an odd prime number appears on the die, a ball is drawn from bag A; otherwise from bag B. If the ball drawn is black, then the probability that it is drawn from bag B is

• A. \(\frac{6}{7}\)
• B. \(\frac{3}{7}\)
• C. \(\frac{1}{5}\)
• D. \(\frac{4}{5}\)

Hint:

When probability asks “given that”, immediately think Bayes theorem and reverse the conditional probability.

Answer 1

The Correct Option is A

Step 1: Decide which die outcomes pick each bag.
An odd prime on a die means $3$ or $5$ (note $2$ is prime but even). So bag $A$ is chosen on 2 of 6 faces, giving $P(A)=\frac26=\frac13$, and bag $B$ on the remaining 4 faces, $P(B)=\frac46=\frac23$.
Step 2: Black-ball chance from bag A.
Bag $A$ has $3+2+2=7$ balls with $2$ black, so $P(\text{black}\mid A)=\frac27$.
Step 3: Black-ball chance from bag B.
Bag $B$ has $1+2+4=7$ balls with $4$ black, so $P(\text{black}\mid B)=\frac47$.
Step 4: Total probability of black.
$P(\text{black})=\frac13\cdot\frac27+\frac23\cdot\frac47=\frac{2}{21}+\frac{8}{21}=\frac{10}{21}$.
Step 5: Apply Bayes' theorem.
$P(B\mid\text{black})=\dfrac{P(B)\,P(\text{black}\mid B)}{P(\text{black})}=\dfrac{\frac23\cdot\frac47}{\frac{10}{21}}=\dfrac{\frac{8}{21}}{\frac{10}{21}}$.
Step 6: Simplify.
This equals $\frac{8}{10}=\frac45$.
\[ \boxed{\dfrac{4}{5}} \]