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There are $20\%$ less white than black balls in a box. Two balls are randomly chosen. If the probability that at least one chosen ball is white is $12/17$, how many black balls are in a box?

If $w$ is the number of white balls, and $b$ is the number of black balls, then: $$w=b-b/5=4b/5$$

Total number of balls in a box is $t=9w/4$ or $t=9b/5$.

How can we find the total number of black balls after two are randomly chosen?

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Hint \begin{align} P( \text{at least } 1 \text{white ball} ) &=1-P(\text{both balls are black}) \\ &=1-\frac{b}{t}\frac{b-1}{t-1} \\ &= 1- \frac59 \left( \frac{b-1}{\frac{9b}{5}-1} \right) \end{align}

equate this to $\frac{12}{17}$, solve for $b$.

  • 0
    How do you get $P$(both black)=$b/t\cdot (b-1)/(t-1)$? – 2017-01-11
  • 1
    I assume the two draws are independent. in the first draw, the probability of getting a black ball would be $\frac{b}{t}$. after which, there are only $t-1$ balls in the box with $b-1$ of them being black. – 2017-01-11