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. $\binom 41\binom {26}2$



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