# Probability of picking numbers

I am doing an introductory course in statistics and I'm struggling with one question in particular.

The question is: Suppose two people each have to select a number from 00 to 99 (therefore 100 possible choices).

(a) The probability that they both pick the number 13 is:

1. 2/100
2. 1/100
3. 1/200
4. 1/10 000
5. 2/10 000

(b) The probability that both persons pick the same number is:

1. 2/100
2. 1/100
3. 1/200
4. 1/10 000
5. 2/10 000

Ok, I understand the (a) part of the question. The probability for the first person to pick 13 is 1/100 and the probability for the second person to pick 13 is also 1/100. Since these are two independent events I use the rule P(A and B) = P(A) * P(B) = 1/10 000.

The (b) part is where I am stuck at the moment. At first, I thought the probability is the same, but instead of only looking for the combination of both people picking 13 we now have to consider the probability of every combination of the same number that the two people can pick.

How do I approach this?

• Can you please tag this as self-study? And, read the self-study tag wiki? – kjetil b halvorsen Apr 28 '17 at 12:24

$\frac{1}{10000}$ is the probability of both people picking a particular number (it could be 13 as you calculated, but any other as well). In how many ways can they pick the same number? Suppose it is $n$, then the probability you require is $\frac{n}{10000}$.
The first person chooses a number. He/she knows what number it will be, so $P(A)=1$. Because the events are still independent, from here you can use the same formula.
All you need is $P(B)$, which is the probability of the second person choosing the same number.