Variance

Question

Comments

C) 7/64 ??

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how ?

E(y) = 11/8 na?

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yes its C..

Explain ...

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see, in atmost 3 tosses if head come it can come only once, since as soon as first head will come event will stop,so

random variable X(# of heads) can take only 2 values i.e 0 or 1.

E(x) = 1*($\frac{1}{2}$ + $\frac{1}{2}.\frac{1}{2}$ + $\frac{1}{2}.\frac{1}{2}.\frac{1}{2}$) = 7/8

E(x²) = 1²*($\frac{1}{2}$ + $\frac{1}{2}.\frac{1}{2}$ + $\frac{1}{2}.\frac{1}{2}.\frac{1}{2}$) = 7/8

var(x) = E(x²) - (E(x))² = 7/8 - 49/64 = 7/64

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i took like 1/2 + 2*1/2² +3*1/2³

which will be used when x = number of toss ryt?

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@Joshi

How r u evaluating E(x²)??

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@Anu007 sir,

but 2 heads or 3 heads can never come, because event will stop as soon as you get first head.

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yes i got that but i ask 

i took like 1/2 + 2*1/2² +3*1/2³

which will be used when x = number of toss ryt  not related to that question.

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E(x²)= $1^{2}\times (\frac{1}{2})+ 1^{2}\times (\frac{1}{2^{2}})+1^{2}\times (\frac{1}{2^{3}})$

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@akash

here x² can also take 2 values i.e 0 or 1, now,

probability to get x²=1, means probability of getting x=1, therfore

E(x²)= 1*P(x=1)

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@Anu sir,

yes exactly.