SOLUTION: A jar contains 3 yellow, 5 red, 4 blue, and 8 green candies. After a candy is selected and not replaced, what is the probability of selecting two red candies?

Algebra ->  Probability-and-statistics -> SOLUTION: A jar contains 3 yellow, 5 red, 4 blue, and 8 green candies. After a candy is selected and not replaced, what is the probability of selecting two red candies?      Log On


   

Question 721354: A jar contains 3 yellow, 5 red, 4 blue, and 8 green candies. After a candy is selected and not replaced, what is the probability of selecting two red candies?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
There are 3+5+4+8 = 20 candies total

P(2 red) = P(Red)*P(Red)

P(2 red) = (5/20)*(4/19)

P(2 red) = (5*4)/(20*19)

P(2 red) = 20/380

P(2 red) = 1/19