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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. $\left(\dfrac{2}{3}\right)$

Sample Space = {HH, TH, HT}
Favorable case ={HH}
=1/3
edited
This question should be tagged under ‘Conditional Probability’ not ‘Binomial Distribution’

in the go book

@Arjun Suresh @Lakshman Patel RJIT please retag this question in conditional probablity

Probability of at least one head, $P_{1H}=\dfrac{3}{4}$

Probability of both heads, $P_{2H}=\dfrac{1}{4}$

Using Bayes' theorem, $P_{2H\mid1H} = \dfrac{P_{2H} \cap P_{1H} }{P_{1H}} =\dfrac{\left(\dfrac{1}{4}\right)}{\left(\dfrac{3}{4}\right)} =\dfrac{1}{3}.$

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??
^^if 1st is head then it will be 1/2
This could also be done by reduced sample space method
yes
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
Sample space:{HH,HT,TH,TT}

Now here

P(HH)=P(TH)=P(HT)=P(TT)=1/4

Now

New Events lets us assume E

E={HH,HT,TH}

Here::

P(HH)=P(TH)=P(HT)=1/3

Therefore

P(HH)=1/3