Question

A bag contains 6 red apples and 8 green apples. You choose 2 apples. What is the probability of picking a red and green apple without replacement?

Answer 1

`24/91`

Explanation:

`"there are a total of "6+8=14" apples"`

`P("red apple")=6/14=3/7`

`"there is no replacement so 13 apples left in bag"`

`P("green apple")=8/13`

`P("red and green apple")=P("red")xxP("green")`

`color(white)(xxxxxxxxxxxxxxx)=3/7xx8/13=24/91`

Answer 2

`48/91`

Explanation:

Let R be the event that a red apple is picked, let G be the event that a green apple is picked.

First pick

`P`(R)`=8/(8+6)=8/14`
`P`(G)`=6/(8+6)=6/14`

Second pick

`P`(G if R first)`=(6)/(14-1)=6/13`

`P`(R if G first)`=(8)/(14-1)=8/13`

So:

`P`(R then G)`=P`(R)`*P`(G if R first)

`=8/14*6/13=48/182`

and

`P`(G then R)`=P`(G)`*P`(R if G first)

`=6/14*8/13=48/182`

`P(RnnG)=48/132+48/182`
`=2(48/182)`

`=48/91`