Question

A bag contain 5 red marbles,4 green marbles and 1 blue marble. A marble is chosen at random from the bag and not replaced; then a second marble is chose. What is the probability both marbles are green

Answer 1

`(2)/(15)`

Explanation:

GIVEN: A bag contain
`5` red marbles,
`4` green marbles and
`1` blue marble. A marble is chosen at random from the bag and not replaced, then a second marble is chose.

TO FIND: What is the probability both marbles are green.

SOLUTION:

Total marbles in bag
`=10`

total number of green marbles
`=4`

Probability that first marble will be green
`P(A)=\frac{\text{total green marbles}}{\text{total marbles}}`

`=(4)/(10)=(2)/(5)`

Probability that second marble will be green
`P(B)=\frac{\text{total green marble in bag}}{\text{total marble in bag}}`

`(3)/(9)=(1)/(3)`

As both events are disjoint

probability both marbles are green
`=P(A)* P(B)`

`=(2)/(5)*(1)/(3)=(2)/(15)`

Hence the probability both marbles are green is
`(2)/(15)`