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den301095 [7]
3 years ago
13

A wallet contains five $10 bills, three $5 bills, six $1 bills, and no larger denominations. If bills are randomly selected one-

by-one from the wallet, what is the probability that at least two bills must be selected to obtain the first $10 bill?
Mathematics
1 answer:
Black_prince [1.1K]3 years ago
6 0

Answer:

\frac{4}{21}

Step-by-step explanation:

Another way of phrasing this question is this; if only two bills are selected without replacement, what is the probability that at one of the two bills will be a $10 bill:

P(One $10 bill) = P(picking a $10 bill first then another type) + (picking another

                            type then a $10 bill)

                        = \frac{5}{15}*\frac{10}{14}+\frac{10}{15} *\frac{5}{14} = \frac{4}{21}

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C. 5x - 2y = 16 is your answer

plug in each point into the equation.

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x = 4, y = 2

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