SOLUTION: Keven had six 10-peso coins and four 5-peso coins. If he took out one coin and the second coin without replacing the first coin, a. What is the probability that both coins were 5-

Question 1016548: Keven had six 10-peso coins and four 5-peso coins. If he took out one coin and the second coin without replacing the first coin,
a. What is the probability that both coins were 5-peso coins?
b. What is the probability that both coins were 10-peso coins?
c. What is the probability that the first coin was a 10-peso coin and the second was a 5-peso coin?

Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
a) (5 peso coin drawn first) * (5 peso coin drawn second)
= (4/10) * (3/9)
= 2/15

b) (10 peso coin drawn first) * (10 peso coin drawn second)
= (6/10) * (5/9)
= 2/3

c) (10 peso coin drawn first) * (5 peso coin drawn second)
= (6/10) * (4/9)
= 4/15
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