Two balls are drawn from a bag containing 5 white and 7 black balls at random. What is the probability that they would be of different colours?

May 30, 2017

Answer:

$\frac{35}{144} + \frac{35}{144} = \frac{70}{144}$

Explanation:

$\text{assuming replacement of the first ball}$

$P \left(\text{white}\right) = \frac{5}{12}$

$P \left(\text{black}\right) = \frac{7}{12}$

$P \left(\text{white and black}\right)$

$= \frac{5}{12} \times \frac{7}{12} = \frac{35}{144}$

$P \left(\text{black and white}\right)$

$= \frac{7}{12} \times \frac{5}{12} = \frac{35}{144}$

Nov 13, 2017

Answer:

If the first ball is replaced: $P \left(\text{different}\right) = \frac{35}{72}$

If the first ball is not replaced: $P \left(\text{different}\right) = \frac{35}{66}$

Explanation:

There are $3$ outcomes that can occur:

• both are white
• both are black
• one of each color

The probability of one of each color is $1 - P \left(\text{same}\right)$

If the first ball is replaced:

$P \left(W W\right) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}$
$P \left(B B\right) = \frac{7}{12} \times \frac{7}{12} = \frac{49}{144}$

:. P("same) = 25/144+49/144 = 74/144

$P \left(\text{different}\right) = 1 - \frac{74}{144} = \frac{70}{144} = \frac{35}{72}$

However, if the first ball is NOT replaced, the probability changes for the second ball. There are fewer of one color and one ball less.

$P \left(W W\right) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}$

$P \left(B B\right) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}$

:. P("same) = 20/132+42/132 = 62/132

$P \left(\text{different}\right) = 1 - \frac{62}{132} = \frac{70}{132} = \frac{35}{66}$