Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of then!/[k!(n−k)!]possible arrangements of the k successes and n-k failures is equally likely.

Bergen

Bergen

Answered question

2021-08-15

Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of the
n!k!(nk)!
possible arrangements of the k successes and n-k failures is equally likely.

Answer & Explanation

Nichole Watt

Nichole Watt

Skilled2021-08-16Added 100 answers

Probability that exactly one of the event E or F occurs is
P(EFcEcF)
Now,
P(EFcEcF)=P(EFc)+P(EcF)P(EFcEcF)
Since P(EFcEcF)=0
P(EFcEcF)=P(EFc)+P(EcF)
Also, P(EFc)=P(E)P(EF) and P(EcF)=P(F)P(EF)P(EFcEcF)=P(E)+P(F)2P(EF)
Hence proved!

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