Payment Probability
Jane carries a credit card which pays in
Bitcoins
. Her card currently holds Bitcoins of two different values:
10
and
5
.
There are 6 coins of value 10 and 4 coins of value 5 (thus, a total of 10 coins). A credit card is allowed to give only one bitcoin at a time. When anyone swipes the credit card, the card picks out a bitcoin out of its pool at random and pays. In total, Jane's card pays 2 bitcoins, one by one.
A vending machine will only accept Jane's payment if the second coin payed is of value 10.
What is the probability that Jane's payment will not be accepted?
Note: Once a bitcoin has been picked to be payed, no new bitcoins replace it in the credit card.
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1 solution
The best method for this is to draw a probability diagram.
Probability of getting the first Bitcoin as value 10 = (6 / 10)
Probability of getting the first Bitcoin as value 5 = (4 / 10)
Probability of getting the second Bitcoin as value 10 = (5/ 9)
Probability of getting the second Bitcoin as value 5 = (3 / 9)
The combinations which will not be accepted are: (10, 5) and (5, 5) - since in these two, the second payment is NOT 10
Therefore, probability of rejection = ((6 / 10) * (4 / 9)) + ((4 / 10) * (3 / 9)) = (36 / 90)
This fraction, simplified, yields 0.4 as the probability. Thus the answer is 0.4