A bag contains 8 red balls and some white b..

Question 1 Report

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.

\frac{1}{3}

\frac{2}{9}

\frac{2}{3}

\frac{8}{33}

Answer Details

Let the number of white balls be n

The number of red balls = 8

Now, the probability of drawing a white ball = $\frac{n}{n + 8}$

The probability of drawing a red ball = $\frac{8}{n + 8}$

Since Pr(White ball) = $\frac{1}{2}$ x Pr(Red ball)

$∴ \frac{n}{n + 8} = \frac{1}{2} \times \frac{8}{n + 8}$

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

$∴ n = 4$

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) = $\frac{8}{12} \times \frac{4}{11}$

$∴$ Pr(Red ball and white ball) = $\frac{8}{33}$

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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 p...

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