Probability that three randomly selected coins will provide enough money to purchase a water bottle You have $8$ coins, 1p, 2p, 5p, 10p, 20p, 50p, £1, £2. Out of these you pick $3$. What is the probability that you will be able to pay for a water bottle that costs 80p? 
I stumbled upon this question and just want to make sure my reasoning is sound. 
If I get a £1 or £2 we can buy the water bottle in all cases so I believe that would be $7C2 + 7C2$ and the only other case in which we could buy the bottle would be if we got the 10, 20 and 50p which is $+1$ case.
So I believe the total probability would be $$\frac{7C2+7C2+1}{8C3}$$ but I feel as if this may be wrong.
 A: To choose $3$ coins with total value of at least $80$p there are three cases:


*

*Highest value coin is £$2$ : $\binom{7}{2} = 21$ possibilities

*Highest value coin is £$1$ : $\binom{6}{2} = 15$ possibilities

*Highest value coin is $50$p : $1$ possibility


There are $\binom{8}{3}=56$ possibilities altogether. Assuming each possibility is equally likely, the probability of choosing $3$ coins with total value of at least $80$p is
$\frac{21+15+1}{56} = \frac{37}{56}$
A: Your listing of the favorable cases is correct. There are ${8\choose3}=56$ equiprobable outcomes. Among these there are ${8\choose3}-{6\choose3}=36$ cases containing at least one of £1 , £2; and there is the special $10+20+50$ pennies case. The probability to draw $\geq80$ pennies therefore is
$${36+1\over 56}\approx0.661\approx{2\over3}\ .$$
A: We correct your count.
You added those cases in which a one pound coin is selected to those in which a two pound coin is selected.  However, that means you count each case in which both are selected twice.  Since we only want to count these cases once, we must them from the total.
A 1 pound coin is selected:  The number of such selections is 
$$\binom{1}{1}\binom{7}{2}$$
as you found since you must select the 1 pound coin and two of the other seven coins.
A 2 pound coin is selected:  The number of such selections is also
$$\binom{1}{1}\binom{7}{2}$$
as you found since you must select the 2 pound coin and two of the other seven coins.
Both a 1 pound coin and a 2 pound coin are selected:  The number of such selections is 
$$\binom{2}{2}\binom{6}{1}$$
since you must select both coins worth at least 1 pound and one of the other six coins.
Therefore, the number of selections that include at least one coin worth at least 1 pound is 
$$\binom{1}{1}\binom{7}{2} + \binom{1}{1}\binom{7}{2} - \binom{2}{2}\binom{6}{1}$$
Since, as you noted, the only other way to obtain at least 80p using three of the eight coins is to select the 10p, 20p, and 50p coins, the number of favorable cases is 
$$\binom{1}{1}\binom{7}{2} + \binom{1}{1}\binom{7}{2} - \binom{2}{2}\binom{6}{1} + \binom{3}{3}$$
Therefore, the probability that you will have enough money to purchase the water bottle when selecting three coins at random is 
$$\frac{\dbinom{1}{1}\dbinom{7}{2} + \dbinom{1}{1}\dbinom{7}{2} - \dbinom{2}{2}\dbinom{6}{1} + \dbinom{3}{3}}{\dbinom{8}{3}}$$
