Answer:
[tex]P(Nickel\ and\ Nickel) = 0.08091[/tex]
Step-by-step explanation:
Given
[tex]Pennies = 18[/tex]
[tex]Dimes = 13[/tex]
[tex]Nickels = 18[/tex]
[tex]Quarters = 13[/tex]
Required
Probability of selecting 2 nickels without replacement
This can be represented as: [tex]P(Nickel\ and\ Nickel)[/tex]
The solution is as follows
[tex]P(Nickel\ and\ Nickel) = \frac{Nickel}{Total}*\frac{Nickel-1}{Total-1}[/tex]
The second nickel is reduced by 1 because the first nickel was not replaced and this also reduces the total by 1
So, we have:
[tex]P(Nickel\ and\ Nickel) = \frac{18}{62}*\frac{18-1}{62-1}[/tex]
[tex]P(Nickel\ and\ Nickel) = \frac{18}{62}*\frac{17}{61}[/tex]
[tex]P(Nickel\ and\ Nickel) = \frac{18*17}{62*61}[/tex]
[tex]P(Nickel\ and\ Nickel) = \frac{306}{3782}[/tex]
[tex]P(Nickel\ and\ Nickel) = 0.08091[/tex]