A box contains 2 red, 5 free, and 3 blue balls. Two balls are drawn randomly. If it is known that neither of the selected balls are blue...

Question

What is the probabiliy that exactly one of the two balls is green?

 

 

 

HELP PLEASE

Answer 1

Use the formula for conditional probability: P(A|B) = P(AB)/P(B)

 

In this case P(exactly one G| no blue) = P(exactly one G AND no blue) / P(no blue)

                                                        = P(GR or RG) / P(no blue)

                                                        = [P(GR) + P(RG)] / P(no blue)

 

                                                         = [(5/10)(2/9) + (2/10)(5/9)] / [(7/10)(6/9)]

 

                                                         = [10/90 + 10/90] / (42/90)

 

                                                         = 20/42

 

If your teacher does not require you to use the conditional probability formula then:

 

P(GR or RG) = P(GR) + P(RG)   (mutually exclusive)

 

                   = (5/7)(2/6)  + (2/7)(5/6) = 20/42

 

Any questions comment back.

Answer 2

Ignore the number of blue balls. They do not figure into the probability since it is known they are not picked. There are 7 possible selections. The possibility of the first one being red is 5/7 5 green to 7 total. The possibility of the second one being green, assuming the first was red, is 5/6 5 green and now 6 total. Since these are independent trials just do 5/6*5/7=25/42 or 59.42%