Probability of winning a raffle

I searched for this answer before posting it, and the threads helped me get as far as I have.

30 tickets are sold in a raffle where 4 prizes will be given. John buys 3 of the tickets. What is the probability that John wins the following:

P(A) = 0 Prizes? , P(B) = 1 prize? P(C) = 2 prizes? P(D) = 3 prizes?

To start off, there are

• $$\dbinom{30}{3}$$ ways of picking three tickets from 30 tickets

• $$\dbinom{3}{3}$$ ways of picking three prizes

• $$\dbinom{3}{2}$$ ways of picking two prizes

• $$\dbinom{3}{1}$$ ways of picking one prize

• $$\dbinom{27}{3}$$ ways of picking zero prizes.

Therefore, $$P(A) = \frac{\dbinom{27}{3}}{\dbinom{30}{3}}$$ $$P(B) = \frac{\dbinom{3}{2}}{\dbinom{30}{3}}$$ $$P(C) = \frac{\dbinom{3}{1}}{\dbinom{30}{3}}$$ $$P(D) = \frac{\dbinom{3}{3}}{\dbinom{30}{3}}$$

$$P(A)=\binom{26}{3}/\binom{30}{3}$$, $$P(B)=\binom{26}{2}\binom{4}{1}/\binom{30}{3}$$, $$P(C)=\binom{26}{1}\binom{4}{2}/\binom{30}{3}$$, $$P(D)=\binom{4}{3}/\binom{30}{3}$$.

This is a fair effort; but in need of improvement.

Notice firstly that you are selecting from 4 prises. So to select three from them would be counted by $$\binom {4}3$$

You have tried to count ways to select zero prises and three non-prises among the three picks; but there are 26 non-prises.   $$\binom {26}3$$

However, likewise you need to count how to select two prises and one non-prise among the three picks. $$\binom 42\binom {26}1$$

Similarly you must count how to select one prise and two non-prises among the three picks. $$\binom 41\binom {26}2$$

And