A student has 5 nickles and 4 quarters. She picks two randomly. What is the expected total value in cents of the coins?

Question

anonymous · Answer

@satellite73 @DanielxAK @mathmate I really need to know this for my midterm tomorrow. Even if it says I'm not viewing the question, please answer. I'll check tomorrow morning before I leave for the test.

anonymous · Answer

@Dido525

anonymous · Answer

Number of ways to choose the Two coins is $$n C k = 9C2 = 36$$

2 nickels can be picked in $$n C k = 5C3 = 10$$ probability = $$\frac{ 10 }{ 36 } =\frac{ 5 }{ 18 }$$

2 quarters can be picked in $$n C k = 4C2 = 6$$ Probability = $$\frac{ 6 }{ 36 }= \frac{ 1 }{6 }$$

1 dime and 1 quarter can be picked in $$1-P(quarters)-p(dimes)= \frac{ 5 }{ 9 }$$

Expected value:$$P(quarter,dime)(0.30)+P(quarter)(0.50)+P(nickels)(0.10)=value$$

=$$(\frac{ 5 }{ 9})(0.30) + (\frac{ 1 }{ 6 })(0.50)+(\frac{ 5 }{ 18 })(0.10) = 27.78=250/9$$ 

in cents.

anonymous · Answer

@mathsss

anonymous · Answer

Thank you!

anonymous · Answer

Welcome :) . Hope your midterm was good :) .