Consider 4 people, each of which has a uniformly random birthday. We ignore leap

Question

Consider 4 people, each of which has a uniformly random birthday. We ignore leap years; thus, one year has 365 days. Define the event A = "at least 2 of these 4 people have the same birthday". What is Pr(A)? (4 2). 1/365 (4 2). 1/365 + (4 3). 1/365^2 + (4 2). 1/365^3 1 - 361. 362. 363. 364/365^4 1 - 362. 363. 364/365^3 Let S = {1, 2, 3, 4, 5, 6, 7}. You choose a uniformly random 3-element subset X of S. Thus, each 3-element subset of S has a probability of 1/(7 3) of being X. Define the event A = "4 is an element of X" What is Pr(A)? 7/15 15/7 3/7 7/3?

Explanation / Answer

1)

At least two people out of 4 i.e.

P(X=2)+P(X=3)+P(X=4)=as mentioned in option B

Therefore, option B is the correct answer.

2)

1C1*6C2/7C3

1*15/35=15/35=3/7= option C

Therefore, option C is the correct answer.

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