# Winning a Lottery (Probability)

John bought 2 tickets to join a lottery, in the lottery they draw 2 tickets out of 100 in the same draw, the owner of the first ticket gets a prize and the owner of the second ticket gets another prize

a- What is the chances of john winning the two prizes b- What is the chances of john not winning anything at all c- What is the chances of john winning one prize

i tried in a- 2/100 and b- 1-0.02 and c- 1/100 is that right ? and if the 2 tickets are not in the same draw will it be the same answers as in one draw ?

• Okay i added it – hassan Dec 13 '15 at 16:38
• Your first result is not right. Suppose your ticket A is drawn firstly. The probability is $\frac{1}{100}$ The probability that your ticket B is drawn secondly is $\frac{1}{99}$ In total $\frac{1}{100}\cdot \frac{1}{99}$. This can happen the other way round. Therefore the final result for a is $2\cdot \frac{1}{100}\cdot \frac{1}{99}$. It doesn´t matter if the tickets are drawn consecutively or not. – callculus Dec 13 '15 at 16:46
• You are welcome. – callculus Dec 13 '15 at 17:58

Drawing $2$ tickets in one draw is the same as drawing $2$ tickets one by one without replacement.

a) John wins two prices if the first ticket drawn is one of his (probability on that is $\frac2{100}$) and - under that condition - also the second ticket drawn is one of his.

That event has probability $\frac2{100}\times\frac1{99}$.

b) For winning no prices both tickets must not be of John.

That event has probability $\frac{98}{100}\frac{97}{99}$.

c) For winning exactly one price there are two possibilities: (1) the first ticket drawn is of John but the second is not; (2) the first ticket drawn is not of John, but the second is.

That event has probability $\frac2{100}\frac{98}{99}+\frac{98}{100}\frac{2}{99}$.

• I understand it now, Thank you very much – hassan Dec 13 '15 at 17:39

Your first result is not right. It doesn´t matter if the tickets are drawn consecutively or not.

Suppose your ticket A is drawn firstly. The probability is $\frac{1}{100}$. The probability that your ticket B is drawn secondly is $\frac{1}{99}$.

In total $\frac{1}{100}\cdot \frac{1}{99}$. This can happen the other way round. Therefore the final result for a is $2\cdot \frac{1}{100}\cdot \frac{1}{99}$.

You can also use the hypergeometric distribution to get the result.

$P(X=2)=\frac{\binom{2}{2}\cdot \binom{98}{0}}{\binom{100}{2}}=\frac{1\cdot 1}{\frac{100\cdot 99}{1\cdot2}}=\frac{2}{100\cdot 99}$

with $X\sim h(N,K,n)=h(100,2,2)$

I hope it helps you to handle b and c.