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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?

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  • $\begingroup$ Can you please tag this as self-study? And, read the self-study tag wiki? $\endgroup$
    – kjetil b halvorsen ♦
    Apr 28, 2017 at 12:24

4 Answers 4

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$\frac{1}{10000}$ 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 $\frac{n}{10000}$.

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  • $\begingroup$ According to the textbook, the probability of (b) happening is 100 * 1/100 = 1/10000, as suggested by Comp_Warrior. It does make sense now, there are n combinations of ways they can pick the same number $\endgroup$
    – Will777
    Aug 23, 2014 at 6:50
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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.

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So...there are a 100 possible choices in the case of selecting a number at random between 0 and 99. The probability of a person picking any number at random is 1/100.Now,the question is what is the probability that both persons pick the same number.Lets call the first person, person1 and the second person2. We know that the probability of each person picking any number at random is 1/100,therefore they both have the same probability individually.The probability of person1's pick does not affect the probability of person2's,therefore they are independent events.For independent events P(A ∩ B) = P(A) * P(B).Therefore the probability that they both pick the same number is 1/100 * 1/100.

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The answer is 1/100. If both people pick a number, the number that person 1(or person 2, it doesn't matter) chooses does not matter. When that number is selected, person 2 has a 1/100 chance of picking the same number, no matter what number person 1 picked.

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    $\begingroup$ The answer to part 1 is 1/10,000. That is 1/100 that the first picks 13 multiplied by 1/100 that the second picks 13 givem the first one did. $\endgroup$
    – Michael R. Chernick
    Jan 9, 2018 at 6:59

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