# MMM10 Solution

THE PROBLEM IS HERE!

In the Quach family, there is an age-old tradition that has been intact since the 1500s. In this tradition, every member of the Quach family receives one lottery ticket for his/her first 25 birthdays.

Based on empirical data collected over the years, it was determined by the family Mathematician that the probability of a Quach family member winning the lottery at least once during that 25 year stretch is 211/243.

Assuming the above statement is true, what is the probability of a Quach family member winning the lottery at least once during any 5 year interval? (Assume that lottery wins are completely independent of each other).

So the probability that a member of the Quach family does not win the lottery in 25 years is 32/243. Since winning each year is independent of any other year, we know:

P(not winning any year)=P(not winning 1st year)*P(not winning 2nd year)*…*P(not winning 25th year)

And since the probability of not winning each year is the same, we get:

P(not winning any year)=P(not winning in nth year)^25
32/243=P(not winning in nth year)^25
P(not winning in nth year)=(32/243)^(1/25)

Now we need to find the probability of winning at least once in a five year interval, which is actually 1-P(not winning in 5 year interval).

P(not winning in 5 year interval)=P(not winning in 1st year)*P(not winning in 2nd year)*…*P(not winning in 5th year)

P(not winning in 5 year interval)=P(not winning in nth year)^5

P(not winning in 5 year interval)=(32/243)^(1/5)=2/3

Since we want to find P(winning at least once in 5 year interval), we know that that will have to be 1/3.

I want me some of them odds.