GATE 2017 - [EE]

Question

A person decides to toss a fair coin repeatedly until he gets a head. He will make at most 3 tosses. Let the random variable Y denote the number of heads. The value of var (Y) where var (.) denotes the variance, equals

(A) 7/8 (B) 49/64 (C) 7/64 (D) 105/64

Comments

@Anu007 how come e(x^2)=e(x)?

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beacuse x= 1 so, x= x²= 1

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For x=3 shouldn’t x^2= 9 while calculating variance   ?

Answer 1

I think its 7/64

Y denotes the no of heads.

Y	0	1
P(Y)	1/8	(1/2)+(1/4)+(1/8)=7/8

E[Y]=0*1/8 + 1*7/8=7/8

E[Y^2]=1^2(7/8);

so Var(Y)=E[Y^2] - (E[Y])^2 = 7/8 - (7/8)^2 = 7/64

Comments

@suvradip everything is correct but P(Y=0) should be 1/8 = P(TTT) as we wont stop on getting single Tail or two Tails. 0 heads can be concluded when we have tried all 3 chances and not received heads. Correct me if I am wrong.

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yes youre correct.