Question

A bag contains 5 balls; 3 balls are drawn in succession

without replacement and two white and 1 black ball are

obtained. If it is known that every ball in the bag is either

white or black, then the probability that, the next draw will

give white ball.

• A. 2/5
• B. 3/5
• C. 1/5
• D. 4/5

Answer 1

Answer: (B)

3/5

Answer 2

Let E₀, E₁, E₂ be the events that the bag contains 2,3,4 white balls.

P(E₀) = P(E₁) = P(E₂) = 1/3.

Let E be the event of drawing 2 white and 1 black balls.

P(E/E₀) = $\frac { { } ^ { 2 } \mathrm { C } _ { 2 } \cdot { } ^ { 3 } \mathrm { C } _ { 1 } } { { } ^ { 5 } \mathrm { C } _ { 3 } } = \frac { 3 } { 10 }$

P(E/E₁) = $\frac { { } ^ { 3 } \mathrm { C } _ { 2 } \cdot { } ^ { 2 } \mathrm { C } _ { 1 } } { { } ^ { 5 } \mathrm { C } _ { 3 } } = \frac { 6 } { 10 }$ P(E/E₂) = $\frac { { } ^ { 4 } \mathrm { C } _ { 2 } \cdot 1 } { { } ^ { 5 } \mathrm { C } _ { 3 } } = \frac { 6 } { 10 }$ .

P(E₀/E) = $\frac { \frac { 1 } { 3 } \times \frac { 3 } { 10 } } { \frac { 1 } { 3 } \times \frac { 3 } { 10 } + \frac { 1 } { 3 } \times \frac { 6 } { 10 } + \frac { 1 } { 3 } \times \frac { 6 } { 10 } } = \frac { 1 } { 1 + 2 + 2 } = \frac { 1 } { 5 }$

P(E₁/E) = 2/5; P(E₂/E) = 2/5.

Probability that next draw will give white ball = P(E₀/E).0 + P(E₁/E) $\frac { 1 } { 2 }$ + P(E₂/E).1

= 0 + $\frac { 2 } { 5 } \times \frac { 1 } { 2 } + \frac { 2 } { 5 } \times 1 = \frac { 3 } { 5 }$ .