You have either a $1-coin or a $2-coin in your right-pants pocket, but you are u

Question

You have either a $1-coin or a $2-coin in your right-pants pocket, but you are unsure which one. You grab a $2-coin from your change container - so you can buy yourself a bottle of water before your statistics class - and put it in your right-pants pocket. Prior to adding the $2-coin to you pocket, you surmise that the chance of you having a $1-coin is 4-times more than likely than having a $2-coin in your right-pants pocket. You arrive at school and order a bottle of water from a food service outlet on campus. The price of the bottle of water is $3.09. Since the only money you have on hand is in your right-pants pocket, you reach into your right-pants pocket and randomly pull out one of the two coins. It is a $2-coin What is the probability that you do not have enough money to purchase the water? Enter your answer to four decimal places.

Explanation / Answer

$4 is enough to buy water, not $3
So, we are being asked P(not enough money to purchase the water)

P(having $1) = 4P(having $2)
Since P(having $1)+P(having $2) = 1
Therefore, 4*P(having $2) + P(having $2) = 1, P(having $2) = .2
P(having $1) = .8

P(having $1)
= P(having 1$|had 2$)
= P(having $1 now and $2 earlier/(having $1 now and $2 earlier + having $2 now and $2 earlier)
= .8*.2/ (.8*.2 + .2*.2)
= .8

So, there is a 80% chance that you don't have enough money to purchase water

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