math

Question

A sequence of independent trails consisting consists flipping of a coin having probability p of coming up of heads is continually performed until either a head occurs or a tital of n flips is made. Let X denote the no of time the coin is flipped.then P(X=n) is

A)(1-p)^n

B)p(1-p)^(n-1)

C)(1-p)^(n-1)

D)np(1-p)^(n-1)

Answer 1

So if head occurs then it will stop , until then (if continuously tails occurs) we are doing this flip .

So if X = 1 then P{H} = p

X=2  P{(T, H)} = (1 − p)p

likewise · · · · · ·

X = n − 1 P{(T, T, . . . , T , H)} = p(1 − p)ⁿ⁻² 

and for X=n then P(X)= p(1 − p)ⁿ⁻¹  

but here they are asking the probability of X=n means when we have chance to flip the coin n times . So if we are having continuously having tails for n-1 times then surely we have chance to flip the coin for n times .

So to have continuous n-1 tails the probability is (1-p)^n-1 . 

Ans is C .

peace

Comments

Ans is C)

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now i fixed it :D .

Answer 2

P(X=n) means we are performing the Trial n times.

 

So after n trials there could Success in the nth trial: in that case P(x=n)=(1-P)^n-1p

and there could be total  faliure : In that case P(X=n) =(1-p)^n

So P(X=n)= Succes+faliure=(1-p)^n-1.