Probability

Probability Level 2

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is 6. Find the probability that it is actually six .

3 8 \dfrac{3}{8} 5 7 \dfrac{5}{7} 6 7 \dfrac{6}{7} 1 2 \dfrac{1}{2}

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Sriven Srilakshmi Pulakaram by Sriven Srilakshmi Pulaka… , 23

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1 solution

let E be the event that the man reports that six occurs in the throwing of the die and s1 be the event that six occurs and s2 be the event that six does not occur.

p(s1)=six occurs=1/6

p(s2)=six doesn't occur =1-1/6=5/6

p(E|s1)=probability that man reports six when six actually occurred =probability that man speaks truth =3/4

P(E|s2)=probability that man reports six when actually did not occur
=probability that man telling lie=1-3/4=1/4

THUS BE BAYES' THEOREM

p(s1|E)=probability that the man is speaking truth

= p(s1)p(E|s1)/ p(s1)p(E|s1)+p(s2)p(E|s2)

=1/6* 3/4 / 1/6 * 3/4 + 5/6 * 1/4 =3/8

1 Helpful 0 Interesting 0 Brilliant 0 Confused

The problem should state "Assume that if the man rolls a 1-5 and lies, he reports a 6." Otherwise, we do not know the probability of him reporting 6 if he rolls 1-5 and decides to lie.

Dan Wilhelm - 5 years, 11 months ago

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