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1. From a bag containing 4 white and 5 black balls a man drawn 3 balls at random. What are the odds against these balls being black?

A. $\frac{5}{37}$

B. $\frac{37}{5}$

C. $\frac{11}{13}$

D. $\frac{13}{37}$

Solution:

Probability of all three balls being black

\begin{aligned} = \frac{^{5} C_{3}}{^{9} C_{3}} \\ = \frac{5}{42} \end{aligned}

Probability that three balls are not black

\begin{aligned} = 1 - \frac{5}{42} \\ = \frac{37}{42} \end{aligned}

Hence, odds against these ball being black

\begin{aligned} = (\frac{37}{42}) : (\frac{5}{42}) \\ = 37 : 5 \end{aligned}

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