# Probability of nth event is between x and y

I have a uniform distribution of age in the range $$[a, b)$$ with $$a=42$$ and $$b=78$$

So the probability that a person walks in a bank that is between $$50$$ and $$70$$ years of age would be $$\frac{70-50}{78-42}$$, right?

But I'm struggling with this problem where I'm supposed to find

1. The probability that 3 out of 7 customers in the bank are between $$50$$ and $$70$$ years of age
2. The probability that the $$7th$$ customer entering the bank is between $$50$$ and $$70$$ years of age
3. The probability that the $$7th$$ customer entering the bank is also the 3rd customer between $$50$$ and $$70$$ years of age

For $$1.$$, is it just $$\frac{20}{\binom{7}{3}}$$?

And for $$2.$$ and $$3.$$, I'm a bit lost so I need some help on them

• You are correct that the probability of any one person in your distribution being between $50$ and $70$ years of age is $5/9$. I suggest starting with Question $2$, since it's the easiest. Then use a binomial distribution to figure out Question $1$ (your answer is not correct). Once you figure out the correct method to solve Question $1$, a slight alteration of that method will set you on the path to solving Question $3$. – Robert Shore Feb 28 at 3:55

Let's assume that the ages of different customers is independent (probably a safe assumption for the purposes of the problem). Let $$p = \frac{70 - 50}{78 - 42}$$.
1. The probability that exactly 3 of the 7 are ages 50-70 is $$\binom{3}{7}p^3(1-p)^4$$. You can view this as a binomial distribution over 7 "coins", each with probability $$p$$ of being heads.
2. The probability of the 7th customer being in the age range is the same as the probability of any other customer being in that age range, which is $$p$$.
• I just re read this and realized that for 1., did you mean $\binom{7}{3}p^3(1-p)^4$ – PTN Mar 1 at 0:42