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 .
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 . Similarly, the probability that the 3rd person has the same birthday is also . So we want to get the Probability that A=B AND A=C: .
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:
: is the probabilty that
all 3 has the same birthday. So is the probability
that at least one person among 3 has the different birthday. This
probability include the cases where , ,
and .
The opposite of the previous question.
Similar idea to the previous question. First let's calculate the probability that all 20 people has different birthday. This probability is . 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: .
There are 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 .
This is basically saying that 2 dice does not match.
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 .
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.
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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 . 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: .