Question

There are 5 blue chips, 4 red chips and 3 yellow chips in a bag. One chip is drawn from the bag. That chip is placed back into the bag, and a second chip is drawn. What is the probability that the two selected chips are of different colors? Express your answer as a common fraction.

Answer 1

Answer:

$\frac{31}{72}$

Step-by-step explanation:

Number of Blue(B) Chips=5

Number of Red(R) Chips=4

Number of Yellow(Y) Chips=3

Total Number of Chips=5+4+3=12

Pr(B)=5/12, Pr(R)=2/12, Pr(Y)=3/12

If two chips of different colors are selected one after the other with replacement, the following combinations are possible.

BR, BY, RB, RY, YB, YR

The probability that the two selected chips are of different colors

Pr(BR OR BY OR RB OR RY OR YB OR YR)=

$=(5/12X2/12) + (5/12 X 3/12) + (2/12 X 5/12) +(2/12 X 3/12) + (3/12 X 5/12) + (3/12 X 2/12)$

=$\frac{10}{144} +\frac{15}{144} +\frac{10}{144} +\frac{6}{144} +\frac{15}{144} + \frac{6}{144}$

=$\frac{62}{144} =\frac{31}{72}$