Probability lotto draw - exactly one number divisible by 9?

Question

What is the probability that a lotto draw (6 out of 45) will draw exactly one number divisible by 9?

I know the probability that it will not draw a number divisible by 9 is $\frac{40}{45} * \frac{39}{44} * \frac{38}{43} * \frac{37}{42} * \frac{36}{41} * \frac{35}{40}$. But how do I calculate the probability for exactly one number divisible by 9?

Answer 1

Hint:

First calculate the probability that the first drawn number is divisible by $9$ and the others are not.

Then multiply this result with ... , any idea of your own here?

Answer 2

Out of the $\binom{45}6$ possible picks, the number of picks where exactly one number is divisble by 9 is the product of

• choosing the number divisible by 9, of which there are 5 of them
• choosing the other 5 numbers in the pick from the 40 numbers not divisible by 9, hence $\binom{40}5$

Thus the desired probability is $\frac{5\binom{40}5}{\binom{45}6}$. This is an example of a hypergeometric distribution.