Question

Dianna has 4 quarters, 3 dimes, and 3 nickels in her purse. She reaches into her purse and randomly grabs two coins, one at a time. What is the probability of first picking a dime and then, without replacing it, selecting a nickel?

Answer 1

At the beginning there are 3 dimes out of 10 coins:
P(first coin is a dime) = 3/10

After that there are 3 nickels out of 9 remaining coins:
P(second coin is a nickel) = 3/9 = 1/3

P(dime, then nickel) = P(first coin is a dime) x P(second coin is a nickel)
= 3/10 x 1/3
= 1/10

Answer:
1/10 . if im wrong im sorry

Answer 2

The probability of first picking a dime and then, without replacing it, selecting a nickel is:

`(1)/(10)`

Explanation:

It is given that:

Dianna has 4 quarters, 3 dimes, and 3 nickels in her purse.

This means that there are a total of: 4+3+3=10 coins

Now, the probability of first picking a dime is: 3/10

and then we will be left with : 9 coins

The probability of selecting a nickel in second choice is: 3/9=1/3

Hence, the total probability is:

`(3)/(10)* (1)/(3)=(1)/(10)`