randy has 4 pennies, 2 nickels , and 3 dimes in his pocket. if he randomly chooses 2 coins, what is the probability that both are dimes?
+26400 Randy has 4 pennies, 2 nickels , and 3 dimes in his pocket. if he randomly chooses 2 coins, what is the probability that both are dimes ?
By the first choice we have 3 dimes of 9 coins = $$\dfrac{3}{9}$$
By the second choice we have 2 dimes of 8 coins = $$\dfrac{2}{8}$$
The probability is $$\dfrac{3}{9}\cdot\dfrac{2}{8} = \dfrac{1}{12} = 0.08\overline{3}$$
+130516 The total outcomes possible = C(9,2) = 36
But, we're just interested in choosing 2 of the 3 dimes = C(3, 2) = 3
So....the probability is 3 / 36 = 1 / 12 = 8.33 %
+26400 Randy has 4 pennies, 2 nickels , and 3 dimes in his pocket. if he randomly chooses 2 coins, what is the probability that both are dimes ?
By the first choice we have 3 dimes of 9 coins = $$\dfrac{3}{9}$$
By the second choice we have 2 dimes of 8 coins = $$\dfrac{2}{8}$$
The probability is $$\dfrac{3}{9}\cdot\dfrac{2}{8} = \dfrac{1}{12} = 0.08\overline{3}$$
