Question
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?
What is the probability that first is white and second is black?
Answer: Option D
The probability that first ball is white:
$$\eqalign{
& = \frac{{^{12}{C_1}}}{{^{30}{C_1}}} \cr
& = \frac{{12}}{{30}} \cr
& = \frac{2}{5} \cr} $$
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
$$\eqalign{
& = \frac{{^{18}{C_1}}}{{^{29}{C_1}}} \cr
& = \frac{{18}}{{29}} \cr} $$
Required probability,
$$\eqalign{
& = \left( {\frac{2}{5}} \right) \times \left( {\frac{{18}}{{29}}} \right) \cr
& = \frac{{36}}{{145}} \cr} $$
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The probability that first ball is white:
$$\eqalign{
& = \frac{{^{12}{C_1}}}{{^{30}{C_1}}} \cr
& = \frac{{12}}{{30}} \cr
& = \frac{2}{5} \cr} $$
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
$$\eqalign{
& = \frac{{^{18}{C_1}}}{{^{29}{C_1}}} \cr
& = \frac{{18}}{{29}} \cr} $$
Required probability,
$$\eqalign{
& = \left( {\frac{2}{5}} \right) \times \left( {\frac{{18}}{{29}}} \right) \cr
& = \frac{{36}}{{145}} \cr} $$
Was this answer helpful ?
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