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Question
```A girl speaks the truth 4 out of 5 times. If she throws a
die and reports it to be a 6, What is the probability of it
being a 6?```
Question Submitted By :: Nagaraju
I also faced this Question!!     Rank Answer Posted By

Re: A girl speaks the truth 4 out of 5 times. If she throws a die and reports it to be a 6, What is the probability of it being a 6?
Answer
# 1
P(A and B)=P(A).P(B)=(4/5).(1/6)=2/15
 Is This Answer Correct ? 6 Yes 4 No
Kaarmukilan S P

Re: A girl speaks the truth 4 out of 5 times. If she throws a die and reports it to be a 6, What is the probability of it being a 6?
Answer
# 2
p(a)=the girl speak truth

p(e)=the dies report

p(e/a)=if the girl speak truth tht the die report is 6

p(a)=4/5
p(e/a)=4/5/1/6

so the probabaility
=p(a)*p(e/a)
=4/5*4/5/1/6
=4/5*24/5
=96/5

 Is This Answer Correct ? 0 Yes 10 No
K Sanghvi

Re: A girl speaks the truth 4 out of 5 times. If she throws a die and reports it to be a 6, What is the probability of it being a 6?
Answer
# 3
The question is "What is the probability of it
being a 6?" The probability is 1 in 6 regardless of what
the girls says.

 Is This Answer Correct ? 2 Yes 2 No
Wareagle

Re: A girl speaks the truth 4 out of 5 times. If she throws a die and reports it to be a 6, What is the probability of it being a 6?
Answer
# 4
A = event that it is actually a six.
B = event that the girl reports it is a six.
required is P[A|B] i.e. Probability that it is actually a six given that the girl reports so.
So, P[A|B] = {P[B|A].P(A)}/ P(B) .....(1)
P[B|A]= P{Girl reports a six| actually a six}
= P{Girl speaks the truth}
= 4/5
P(A)=P{ it is a six }= 1/6
P(B)=P{Girl reports a six}
=P{Girl reports a six, A}+P{Girl reports a six, it is not a six}
=P[B|A].P(A)+P[B|A'].P(A')
=(4/5).(1/6)+(1/5).(5/6)
= 3/10
Hence, from equation (1)
P[A|B]= {(4/5)(1/6)}/(3/10) = 4/9

 Is This Answer Correct ? 1 Yes 0 No
Dijit

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