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?

Question

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.

Answer 1

To getting at least 80 cents:

we have 1€ and two from the rest: ${7\choose 2} =21$

we don't have 1€, but we have 2€: ${6\choose 2} =15$

we don't have 1€ and not 2€, then it is not possible if we have 10,20 and 50 cents

So we have 37 good possibilites among ${8\choose 3} = 56$ so the probability is $$P= {37\over 56}$$

Answer 2

More elegant (IMO) and easily doable with paper and pen:

there are $~2~$ ways to get at least $~80c~$

$~1: 50+20+10 \to$ probability is $~3/8~$ for first draw, $~2/7~$ for $2$nd draw, $~1/6~$ for third draw for total of: $$(3*2*1)/(8*7*6)$$

$~2: 100 +~$ anything or $~ 200+$anything

which is $~1-(~$Probability of not getting $~100~$ or $~200)~$

Probability of not getting $~100~$ or $~200~$ is $~6/8~$ for first draw, $~5/8 ~$ for $2$nd draw and $~4/8~$ for $3$rd draw

so total probability for option $~2~$ is $$1-(6*5*4)/(8*7*6)$$

So total probability to get at least $~80c~$ is

$$3*2*1/(8*7*6)+1-(6*5*4)/(8*7*6)$$ Simplify the fractions by dividing out the $6:$ $$1/(8*7)+1-5*4/(8*7)$$ $$=1+1/56-20/56$$ $$=(56+1-20)/56$$ $$=37/56$$