Andy has 6 dimes and 10 nickels in his pocket. If he draws two coins without replacing the first coin, what is the probability of drawing a nickel and then a dime?
A ¼
B 15/64
C 1/60
D 3/5
prob(nickel, then dime)
= (10/16)(6/15)
= 1/4
To find the probability of drawing a nickel and then a dime, you need to determine the number of favorable outcomes (drawing a nickel and then a dime) divided by the total number of possible outcomes.
First, let's find the total number of possible outcomes. Andy has 16 coins in total (6 dimes + 10 nickels). When he draws the first coin, he has 16 options. After selecting the first coin, there are 15 coins left in his pocket.
To draw a nickel first, Andy has 10 options (since there are 10 nickels). After selecting a nickel, there are now 5 dimes remaining in his pocket.
To draw a dime second, Andy has 5 options (since there are 5 remaining dimes).
So the total number of favorable outcomes (nickel then dime) is 10 (options for drawing the nickel) multiplied by 5 (options for drawing the dime) = 50.
Therefore, the probability of drawing a nickel and then a dime is 50 (favorable outcomes) divided by 16 (total possible outcomes) multiplied by 15 (remaining coins after drawing the first coin).
50 / (16 * 15) simplifies to 50 / 240, which reduces to 5 / 24.
Therefore, the correct answer is not listed among the options.