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A bag contains 8 red balls and some white balls. If the p...

A bag contains 8 red balls and some white balls. If the probability of drawing a white ball is half of the probability of drawing a red ball then find the probability of drawing a red ball and a white ball if the balls are drawn without replacement.

A.

13

B.

29

C.

23

D.

833

Correct answer is D

Let the number of white balls be n

The number of red balls = 8

Now, the probability of drawing a white ball = nn+8

The probability of drawing a red ball = 8n+8

Since Pr(White ball) = 12 x Pr(Red ball)

= \frac {n}{n + 8} = \frac {4}{n + 8}

\therefore n = 4

Number of white balls = 4

The possible number of outcomes = n + 8 = 4 + 8 = 12

Pr(Red ball and White ball) = Pr(Red ball) x Pr(White ball)

Pr(Red ball) = \frac {8}{12}

Pr(Red ball) = \frac {4}{11}

Pr(Red ball and white ball) = \frac {8}{12} \times \frac {4}{11}

\thereforePr(Red ball and white ball) = \frac {8}{33}