Question

A bag has 14 marbles, all identical except for their color. There are 7 red marbles, 5 blue marbles,and 2 yellow marbles.If you draw two marbles from the bag WITHOUT replacement, find the following probabilities:**LEAVE ALL ANSWERS AS FRACTIONSa.) P(RR)b.) P(RB) =c.) P(RY) -d.) P(BR) =e.) P(BB)f.) P(BY) -g.) P(YR) =h.) P(YB) =l) P(YY)

Answer 1

SOLUTION:

Step 1:

A bag has 14 marbles, all identical except for their color.

There are 7 red marbles, 5 blue marbles, and 2 yellow marbles.

If you draw two marbles from the bag WITHOUT replacement,

find the following probabilities:

7 red marbles , 5 blue marbles and 2 yellow marbles

Total = 7 + 5 + 2 = 14

( Without Replacement)

a.) P(RR) =

$\frac{7}{14}\text{ x }\frac{6}{13}=\frac{42}{182}=\frac{3}{13}$$P(R\R)\text{ = }\frac{3}{13}$

b.) P(RB) =

$\frac{7}{14}X\text{ }\frac{5}{13}=\frac{35}{182}=\frac{5}{26}$$P\text{ ( RB) =}\frac{5}{26}$

c.) P(RY) =

$\frac{7}{14}X\frac{2}{13}=\frac{14}{182}=\frac{1}{13}$$P(RY)\text{ =}\frac{1}{13}$

d.) P(BR) =

$\frac{5}{14}X\frac{7}{13}=\frac{35}{182}=\frac{5}{26}$$P(BR)\text{ =}\frac{5}{26}$

e.) P(BB) =

$\frac{5}{14}\text{ X }\frac{4}{13}=\frac{20}{182}=\frac{10}{91}$$P(BB\text{ ) =}\frac{10}{91}$

f.) P(BY) =

$\frac{5}{14}X\text{ }\frac{2}{13}=\frac{10}{182}=\frac{5}{91}$$P(BY)\text{ =}\frac{5}{91}$

g.) P(YR) =

$\frac{2}{14}\text{ X}\frac{7}{13}=\frac{14}{182}=\frac{1}{13}$$P(YR)\text{ =}\frac{1}{13}$

h.) P(YB) =

$\frac{2}{14}\text{ X}\frac{5}{13}=\frac{10}{182}=\frac{5}{91}$$P(YR)\text{ = }\frac{5}{91}$

l) P(YY) =

$\frac{2}{14}\text{ x}\frac{1}{13}=\frac{2}{182}=\frac{1}{91}$$P(YY)\text{ =}\frac{1}{91}$