Customers arrive at a store at the rate of 10 per hour. Each is either male or
female with probability 1/2. Assume that you know that exactly 10 women entered within some
hour (say, 10 to 11am). (a) Compute the probability that exactly 10 men also entered. (b)Compute the probability that at least 20 customers have entered.
Part (a) is clear to me and its answewr is:
$$\frac{\exp(-5) (5)^{10}}{10!}$$,
I am just confused in the second part, i.e., (b) Compute the probability that at least 20 customers have entered.
The example is 18.2 and is taken from here
I believe part (b)'s solution as:
$$1-\sum_{n=0}^{19} \frac{\exp(-10) (10)^n}{n!}$$, however, the answer is given as:
$$1-\sum_{n=0}^9 \frac{\exp(-5) (5)^n}{n!}$$
Can anybody explain, also I am unable to understand the significance of giving this line:
Assume that you know that exactly 10 women entered within some hour (say, 10 to 11am)
How does it make difference to the answer of part (b)?
w
Best Answer
Let $X$ be the total number of people who arrived in the given hour. We can divide this number into the number of men (denoted $M$) and the number of women (denoted $W$). Clearly, $X=M+W$.
We want to calculate the probability that $X \geq k$, where $k=20$ in this problem. In other words, we want to know $\text{P} (X \geq k)$. However, we already know that $W=a$, where $a=10$ in this problem. So we have:
$$ \text{P} (X \geq k)=\text{P} (M+W \geq k)=\text{P} ( M \geq k-a) $$
This makes the problem an issue of conditional probability, which is why the number of women makes a difference in the solution.
Another important piece is that that total arrival rate of people is $\lambda=10$ people per hour. But the arrival rate of men is $\lambda_m=\lambda/2=5$ men per hour (equivalent for women).
Then, we can calculate:
$$ \text{P} (M \geq k-a)=\text{P} (M > k-a) + \text{P} (M=k-a)=1-\text{P} (M \leq k-a) + \text{P} (k-a) $$
where we performed the manipulations above so we can easily apply the cumulative distribution function of the Poisson distribution, which gives:
$$ \text{P} (M \geq k-a)=1- \left( e^{-\lambda_m} \sum_{i=0}^{\lfloor k-a \rfloor} \frac{\lambda_m^i}{i!} \right) + \frac{e^{-\lambda_m} \lambda_m^{k-a}}{(k-a)!}=1- \left( e^{-\lambda_m} \sum_{i=0}^{\lfloor k-a-1 \rfloor} \frac{\lambda_m^i}{i!} \right) $$
Plugging in our known parameters:
$$ \text{P} (M \geq 10)=1-e^{-5} \sum_{i=0}^9 \frac{5^i}{i!} $$
which is the same solution as the given answer.