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

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) =

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

b.) P(RB) =

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

c.) P(RY) =

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

d.) P(BR) =

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

e.) P(BB) =

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

f.) P(BY) =

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

g.) P(YR) =

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

h.) P(YB) =

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

l) P(YY) =

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