# 15 Coins, 7 are dimes; What is the probability of picking at least 1 dime in 5 picks?

Cecil has a $$15$$ coin collection:

• Four coins are quarters, seven coins are dimes, three are nickels and one is a penny.
• For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins.

So $$1$$ pick out of the $$5$$ has to be a dime at the very least, but overall there are $$15$$ coins and $$7/15$$ are dimes. I have no clue how to set this up.

Edit: (in response to the comments)

Oh so picking no dime would be anything other than the $$7$$ so $$8/15$$.. But there are $$\binom{15}{5}$$ ways of selecting $$5/15$$ available coins and $$7$$ of those are dimes.

What do I do now $$?$$.

• "So 1 pick out of the 5 has to be a dime at the very least" - that is not true. We could for instance pick four quarters and one nickel. Commented Oct 12, 2016 at 7:46
• Hint on question in title: find the probability of picking no dimes at all in 5 picks. Commented Oct 12, 2016 at 7:47
• Hint: Try to compute the probability of choosing no dime. Commented Oct 12, 2016 at 7:47
• Oh so picking no dime would be anything other than the 7 so 8/15.. But there are (15 5) ways of selecting 5/15 available coins and 7 of those are dimes. What do I do now? Commented Oct 12, 2016 at 8:00

As you noted in the comments, there are $$\binom{15}{5}$$ ways to select five of the fifteen coins. Since there are seven dimes, there are eight coins that are not dimes. The number of ways we could select five of these eight coins is $$\binom{8}{5}$$ Therefore, the probability of selecting no dimes in the five picks is $$\frac{\dbinom{8}{5}}{\dbinom{15}{5}}$$ To find the desired probability, subtract this number from $1$.