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.
13
29
23
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}
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