probability

Question

a fair coin is tossed 10 times ,wt is the probability that head will come up in first two toss of coin??

Answer 1

"a fair coin is tossed 10 times ,wt is the probability that head will come up in first two toss of coin"

We do not even need to consider remaining tosses.

So Total Set of outcomes of First 2 tosses is = { HH,HT,TH,TT}

Required outcome = {HH}

Answer -> cardinality {HH} / cardinality {HH,HT,TH,TT}  = 1/4 = 0.25

Comments

this example can be solve using binomial theorem .

how ?

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HI tiger binomial will not can not apply on for first two pass condition.

Answer 2

No of way(s) to fillFirst two places = 1

Rest tosses may result Head or Tail = 2*2*2*.......2 = 2^8 = 256

Total sample space = 2*2*2.....*2 = 2^10 = 1024

P (1st two tossesresult Head) = 256/1024 = 1/4 = 0.25

Comments

if the question was like that
" a fair coin tossed 10 times ,wt is the probability that only the first two tosses will yield head?"
is this question and above are same na.??

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"probability that first two tosses will yield head?" Means first two tosses turns Head rest 8 May be head may be Tail. P = 1/4

"Probability that only the first two tosses will yield head?" Means Head on first two and Tail on last 8 Coins. P = 1/1024

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yes got now. :) thanks

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Though answer is correct, we can also do by considering first 2 cases only- probability of head in first attempt = 0.5, in second attempt = 0.5. Since these two events are independent required probability = 0.5 * 0.5 = 0.25.

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Can't it be solve with binomial???????

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@Arjun sir

Cant we apply bionomial  when it is given that WHAT IS THE PROBABILITY THAT ONLY THE FIRST TWO TOSSES WILL YIELD HEADS?pls answer

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binomial will not can not apply on for first two pass condition.