Question

Box A contains 3 white and 2 black balls and box B contains 2w and 3B balls. If a folded person selects a ball from one of the boxes at random, then the probability that it is white let A and B be the events of selecting the first and second boxes respectively.

• A. 1/3
• B. ¼
• C. 1/5
• D. ½

Answer 1

Answer: (D)

½

Answer 2

P(1) = $\frac { 1 } { 2 }$; P(2) = $\frac { 1 } { 2 }$ .

Let E be the event of selecting white ball.

P(E/A) = $\frac { 3 } { 5 }$; P(E/B) = $\frac { 2 } { 5 }$.

P(A∩E) = P(1). P(E/A) = $\frac { 3 } { 10 }$.

P(B∩E) = P(2) P(E/B) = $\frac { 1 } { 5 }$

Total probability of E = P(5) = P(A∩E) + P(B∩E)

= $\frac { 3 } { 10 } + \frac { 1 } { 5 } = \frac { 1 } { 2 }$ .