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1. A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement.
What is the probability that first is white and second is black?

A. $\frac{18}{145}$

B. $\frac{18}{29}$

C. $\frac{36}{135}$

D. $\frac{36}{145}$

Solution:

The probability that first ball is white:

\begin{aligned} = \frac{^{12} C_{1}}{^{30} C_{1}} \\ = \frac{12}{30} \\ = \frac{2}{5} \end{aligned}

Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:

\begin{aligned} = \frac{^{18} C_{1}}{^{29} C_{1}} \\ = \frac{18}{29} \end{aligned}

Required probability,

\begin{aligned} = (\frac{2}{5}) \times (\frac{18}{29}) \\ = \frac{36}{145} \end{aligned}

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