In a raffle they sell $30$ tickets and offer $7$ prizes. They place the tickets in a hat and draw one for each prize.

a) The tickets are sampled without replacement, i.e. the selected tickets are not placed back in the hat. What is the probability of winning one prize if you buy one ticket?

b) What if the tickets are sampled with replacement?

In the answers for the exercise the probability of not winning is calculated first.

If I wanted to answer the questions by calculating the probability of winning first, how would I do that?

For a) I tried $(7/30)(6/29)(5/28)(4/27)(3/26)(2/25)(1/24)$. Why is this wrong? How would you do it instead?

Thanks in advance


1 Answer 1


There are many ways to approach these problems and different ways to think about it depending on how comfortable you are with each.

For when tickets are drawn with replacement, imagine briefly that we go ahead and draw each ticket, one after another, even beyond the normal stopping point. We have drawn all thirty tickets. Now... the position in the line of tickets that your ticket is at is equally likely to be one position to another. There are seven "good" places for your ticket to have come out out of $30$, so your probability is very simply:


Another way to approach, to win is equivalent to having not lost. If you lose, this means that the first ticket chosen is not yours, neither is the second ticket, nor the third, etc... Subtracting this away from $1$ gives the probability of you having won something which is:

$$1 - \dfrac{29}{30}\times\dfrac{28}{29}\times\dfrac{27}{28}\times\dots\times\dfrac{23}{24} = 1-\dfrac{23}{30}=\dfrac{7}{30}$$

Yet another which is similar to the previous is where we draw tickets simultaneously rather than in sequence, letting us use binomial coefficients. We get:

$$1 - \dfrac{\binom{29}{7}}{\binom{30}{7}} = \dfrac{7}{30}$$

The probability that you calculated of $\frac{7}{30}\times\frac{6}{29}\times\dots\times\frac{1}{24}$ is a correct answer to the incorrect problem. This is the probability that you and your six friends who each purchased one ticket a piece all won a prize at this raffle.

Now... on to the problem of if there was replacement and a person could win multiple times with a single ticket... the probability of you winning at least once is again $1$ minus the probability of not winning anything at all. Not winning anything on the first occurs with probability $\frac{29}{30}$, just as it is on the second and third, etc... and these events are assumed to be independent, giving the probability of winning at least once when tickets are drawn with replacement as being:

$$1 - \left(\dfrac{29}{30}\right)^7$$

If you were instead interested in the probability of winning exactly one prize (so we don't count if we won two prizes or more, nor do we count if we didn't win anything) then we can use the binomial distribution here. Pick which prize specifically you won, then multiply by the probabilities of actually having won that prize and having not won the others, giving:

$$7\times \left(\dfrac{29}{30}\right)^6\times \dfrac{1}{30}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .