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24 votes
24 votes

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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3 Comments

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

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

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4 Answers

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33 votes
Best answer

Answer - A

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}.$

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4 Comments

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
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This could also be done by reduced sample space method
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yes
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21 votes
21 votes
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
6 votes
6 votes
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
1 vote
1 vote
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
Answer:

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