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If two fair coins are flipped and at least one of the outcomes is known to be a head, what is the probability that both outcomes are heads?

1. $\left(\dfrac{1}{3}\right)$
2. $\left(\dfrac{1}{4}\right)$
3. $\left(\dfrac{1}{2}\right)$
4. $\left(\dfrac{2}{3}\right)$
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Sample Space = {HH, TH, HT}
Favorable case ={HH}
=1/3

prob(at least one head) = $\dfrac{3}{4}$

prob(both heads) =$\dfrac{1}{4}$

using bayes' theorem =$\dfrac{\left(\dfrac{1}{4}\right)}{\left(\dfrac{3}{4}\right)} =\dfrac{1}{3}.$

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I did small mistake, I took it first case is head ad then what is the outcome!

In this case, answer can be 1/4 na??
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^^if 1st is head then it will be 1/2
When two coins are flipped , sample space is { HH, HT, TH, TT }

Now if it is known that at least one of them is head then, reduced sample space will be --> { HH, HT, TH}

In this reduced sample space only one sample point fulfilling the criteria( both are heads HH) hence required probability is 1/3
Using conditional probabilty

P(A/B) = P(A∩B)/P(B)

P(A∩B) =  1/4

P(B) = 3/4

the probability is 1/4/(3/4)=1/3

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