Answers to the Practice of probability

1. 2 people have the same birthday. Assume 365 days/year, birthdays randomly distributed over a year.

   The 1st person can choose a whatever birthday (e.g., say April 1). Then the probability that the 2nd person has the same birth day (April 1) is .

2. 3 people have the same birthday.

   Let's name the 3 people: A, B and C. The 1st person (A) can choose a whatever birthday (e.g., say April 1). Then the probability that the 2nd person has the same birthday (April 1) is $Pr(A=B) = 1/365$. Similarly, the probability that the 3rd person has the same birthday is also $Pr(A=C) = 1/365$. So we want to get the Probability that A=B AND A=C: .

3. All 3 people have the different birthday.

   The 1st person (A) can choose a whatever birthday (e.g., April 1). B can choose any birthday other than April 1, so the probability is (365-1)/365. Let's say B's birthday is March 1. C can choose any day other than March 1 and April 1, so the probability is (365-1-1)/365.

   The correct answer is .

   Common mistake:
   $1 - (1/365)^2$: $(1/365)^2$ is the probabilty that all 3 has the same birthday. So $1 - (1/365)^2$ is the probability that at least one person among 3 has the different birthday. This probability include the cases where $A = B \not= C$, $A \not= B = C$, and $A = C \not= B$.

4. In a group of 3 people, at least one pair has the same birthday.

   The opposite of the previous question.

5. In a group of 20 people, at least one pair has the same birthday.

   Similar idea to the previous question. First let's calculate the probability that all 20 people has different birthday. This probability is $\frac{364}{365} \cdot \frac{363}{365} \cdot \frac{362}{365} \cdots \frac{346}{365} = \frac{365!}{(365-20)! (365)^{20}} = 0.588562$. So is the probability that at least one pair has the same birthday in a group of 20 people. It's a pretty high probability, isn't it?

   To generalize this, the probability that at least one pair has the same birthday in a group of n people is: $1 - \frac{365!}{(365-n)! 365^n}$.

6. Throw 2 dice simultaneously. What is the probability that the product of these 2 values are 6?

   There are $6 \cdot 6 = 36$ elementary outcomes: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), ...., (6,6). Of these elementary outcomes , only (1,6), (2,3), (3,2) and (6,3) can give the event that the product of these 2 values to be 6. The answer is .

7. Throw 2 dice simultaneously. Probability that one value is greater than the other.

   This is basically saying that 2 dice does not match.

8. If two dice are rolled three times, what is the probability that the two dice will match (i.e., display the same number) on one of the three rolls?

   From one roll of two dice, the probability that the two dice match is 1/6, and the probability that the two dice do not match is $1 - 1/6 = 5/6$.

   This problem can be handled by binominal distribution if you define that R.V. X is the number of times that 2 dice match. Then we can use n=3, and the probabilitty of success can be defined as the probability that 2 dies match: p=1/6.

   $Pr[X=k] = {n \choose k} p^k (1-p)^{n-k} = {3 \choose k} (1/6)^k (5/6)^{n-k}$.

   So the question asks the probability that X=1: .

   If the question were to obtain the probability that the two dice will match on at least one of the three rolls we would get $Pr[X \ge 1] = Pr[X=1] + Pr[X=2] + Pr[X=3]$. An easier way to calculate this probability is to subtract the probability that the two dice doesn't match in any of the 3 trials from 1: $1 - Pr[X=0] = 1 - {3 \choose 0} (1/6)^0 (5/6)^{3 - 0} = 1-(5/6)^3 = 0.4212963$.