There is a machine with 8 toys in it that each cost between 25 cents and 2 dollars, with each toy being 25 cents more expensive than the next most expensive one. Each time Sam presses the big red button on the machine, the machine randomly selects one of the remaining toys and gives Sam the option to buy it. If Sam has enough money, he will buy the toy, the red button will light up again, and he can repeat the process. If Sam has 8 quarters and a ten dollar bill and the machine only accepts quarters, what is the probability that Sam has to get change for the 10 dollar bill before he can buy his favorite toy- the one that costs $1.75? Express your answer as a common fraction.

I am stuck on this one... I keep getting it wrong, so any help?

everythinguniversal9 Jun 7, 2019

#1**+1 **

I am sorry that I couldn't finish the problem. The best method I can think of for this problem is listing the cases

Try to find all the cases of him buying a toy. The number of those cases will be the denominator of your answer

Then find all the cases of him getting change for the 10 dollar bill before him buying his favorite toy. The number of those cases will be the numerator of your answer.

This might be a little tedious! I hope this helps!

CalculatorUser Jun 7, 2019

#2**0 **

Thanks for your explanation, but I'm looking for a faster way to do it! Thanks for the help!

😏😏

everythinguniversal9
Jun 7, 2019

#3**+2 **

Here's my best attempt....I'm not good at these, so...I can't guarantee that I'm correct....

He will not need change if

(1) He selects the 1.75 toy first ....or....

(2) He selects the .25 toy first and the 1.75 toy second

The probability of the first = 1/8

And the probability of the second is (1/8) (1/7) = 1/56

So...the probability that he will need change is 1 - P(that he won't need change ) =

1 - ( 1/8 + 1/56) =

1 - ( 8/56) =

56/56 - 8/56 =

48/56 =

6/7

CPhill Jun 8, 2019