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...

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.

Answer Details

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 $\frac{\mathrm{<br>}}{}$

Since Pr(White ball) = 12 $\frac{1}{2}$ x Pr(Red ball)

∴nn+8=12×8n+8 $<br>$

= nn+8=4n+8 $\frac{\mathrm{<br>}}{}$

∴n=4 $<br>$

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) = 812

Pr(Red ball) = 411 $\frac{\mathrm{<br>}}{}$

Pr(Red ball and white ball) = 812×411 $\frac{\mathrm{<br>}}{}$

∴ $\therefore$Pr(Red ball and white ball) = 833