# What's the probability that picking $3$ coins from a bag containing all coins from $1$ cent through $2$ euro will yield a total of $80$ cents?

I recently ran into this interview question and am wondering if my solution is correct. The setting is

We have a bag containing one coin of each type, i.e. we have one $$1$$ cent coin, one $$2$$ cent coin, one $$5$$ cent coin, one $$10$$ cent coin, one $$20$$ cent coin, one $$50$$ cent coin, one $$100$$ cent coin and one $$200$$ cent coin. We random select three coins from this bag, what is the probability that the sum of the values of the coins is at least $$80$$ cents? Start off with giving an initial guess for this probability, without computing anything, and explain your estimate.

My solution: For the estimation part, I figured there were $$\binom{7}{2}$$ configurations of three coins that contain the $$200$$ cent coin. Furthermore, there's also $$\binom{7}{2}$$ configurations of three coins that contain the $$100$$ cent coin. This already sums to $$21+21$$ configurations. However, this sum contains some configurations twice, so it is an overestimation. In total there are $$\binom{8}{3}=56$$ configurations of three coins. Thus the probability of taking three coins with a total value of atleast $$80$$ cents, is approximately $$\frac{42}{56}\approx\frac{2}{3}$$ (rough estimation, I know).

For the exact solution. I know that there are $$\binom{7}{2}$$ configurations of three coins that contain the $$200$$ cent coin, thus automatically satisfying that the sum is above $$80$$ cents. Then, to avoid duplicate configurations, there are $$\binom{6}{2}$$ configurations of coins containing the $$100$$ cent coin. Then there is one last configuration of $$50$$ cents, $$10$$ cents and $$20$$ cents, that also makes $$80$$ cents. Thus there are $$\binom{7}{2}+\binom{6}{2}+1=21+15+1=37$$ configurations. Thus the probability of taking atleast $$80$$ cents when randomly grabbing $$3$$ coins, is $$\frac{37}{56}$$.

Is this correct? Any help is appreciated.

• You really should list the values of all the coins explicitly. Not everybody lives in the Euro zone! – TonyK Feb 16 at 21:20
• For anyone curious, the euro coin denominations are 1 cent, 2 cents, 5 cents, 10 cents, 20 cents, 50 cents, 1 euro and 2 euro. (And 1 euro is 100 cents of course.) – Minus One-Twelfth Feb 16 at 21:23
• So it was sheer laziness? – TonyK Feb 16 at 21:37
• @TonyK I edited the post. Hope you can feel relaxed once more. – S. Crim Feb 16 at 21:41
• @TonyK Fair enough, sorry. – S. Crim Feb 16 at 22:00

we have 1€ and two from the rest: $${7\choose 2} =21$$
we don't have 1€, but we have 2€: $${6\choose 2} =15$$
So we have 37 good possibilites among $${8\choose 3} = 56$$ so the probability is $$P= {37\over 56}$$
• That is incorrect for multiple reasons, but also because there are $8$ different coins in total. – S. Crim Feb 16 at 21:03