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?

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Answer 1 · 2017-05-30T18:47:16

$35/144+35/144 =70/144$

$"assuming replacement of the first ball"$

$P("white")=5/12$

$P("black")=7/12$

$P("white and black")$

$=5/12xx7/12=35/144$

$P("black and white")$

$=7/12 xx5/12 = 35/144$

Answer 2 · 2017-11-13T07:31:08

If the first ball is replaced: $P("different") = 35/72$

If the first ball is not replaced: $P("different") = 35/66$

There are $3$ outcomes that can occur:

The probability of one of each color is $1- P("same")$

If the first ball is replaced:

$P(WW) = 5/12 xx5/12 = 25/144$
$P(BB) = 7/12 xx7/12 = 49/144$

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

$P("different") = 1-74/144 = 70/144 = 35/72$

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.

$P(WW) = 5/12 xx4/11 = 20/132$

$P(BB) = 7/12 xx6/11 = 42/132$

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

$P("different") = 1-62/132 = 70/132 = 35/66$