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?

2 Answers
Jim G.
Nov 8, 2017

#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#

Tom M.
Nov 8, 2017

#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#

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