Problem Solving

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

1/8
2/8
3/4
3/8
1

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

1/8
2/8
3/4
3/8
1

Let T be event he tells truth

L event he tells a lie

S be event it is a 6 S^c event it is not a 6

P(T) = 3/4


We want to find the probability he is telling the truth. By total probability

P(T) = P(T|S) P(S) + P(T|S^c) P(S^c)
= 3/4 x 1/6 + 0 (If it is not a 6 then he cannot be telling the truth so that probability of truth telling is zero)
= 1/8

Choose A

dtweah wrote:

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

1/8
2/8
3/4
3/8
1

Let T be event he tells truth

L event he tells a lie

S be event it is a 6 S^c event it is not a 6

P(T) = 3/4


We want to find the probability he is telling the truth. By total probability

P(T) = P(T|S) P(S) + P(T|S^c) P(S^c)
= 3/4 x 1/6 + 0 (If it is not a 6 then he cannot be telling the truth so that probability of truth telling is zero)
= 1/8

Choose A

It is miserable to go wrong even after using Baye's theorem as done here. I was thinking somebody will do it in non bayes ways. Baye's approach is as below:

P(A) = Prob that he reports a six
P(E) = He speaks truth = 3/4
P(E') = He speaks false = 1/4

P(F) = get six in a through = 1/6
P(F') = do not get a six = 5/6

He reports 6 in two ways when he speaks truth and it is indeed a 6 or it is not a six and he speaks false:

p(A) = P(E)*P(F) + P(E')*P(F') = 3/4*1/6 + 1/4*5/6 = 8/24 = 1/3

So it is actually a 6 is : P(E)*P(F)/P(A) = 3*3/24 = 3/8

maihuna wrote:

dtweah wrote:

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

1/8
2/8
3/4
3/8
1

Let T be event he tells truth

L event he tells a lie

S be event it is a 6 S^c event it is not a 6

P(T) = 3/4


We want to find the probability he is telling the truth. By total probability

P(T) = P(T|S) P(S) + P(T|S^c) P(S^c)
= 3/4 x 1/6 + 0 (If it is not a 6 then he cannot be telling the truth so that probability of truth telling is zero)
= 1/8

Choose A

It is miserable to go wrong even after using Baye's theorem as done here. I was thinking somebody will do it in non bayes ways. Baye's approach is as below:

P(A) = Prob that he reports a six
P(E) = He speaks truth = 3/4
P(E') = He speaks false = 1/4

P(F) = get six in a through = 1/6
P(F') = do not get a six = 5/6

He reports 6 in two ways when he speaks truth and it is indeed a 6 or it is not a six and he speaks false:

p(A) = P(E)*P(F) + P(E')*P(F') = 3/4*1/6 + 1/4*5/6 = 8/24 = 1/3

So it is actually a 6 is : P(E)*P(F)/P(A) = 3*3/24 = 3/8

MISERABLE?? Hmm. I am thinking it is you who are definitely feeling that way. Keep copying solutions and misleading yourself into thinking you are solving them. It will show in the end. Keep bringing it on.

hi

the way maihuna has calculated the porb using bayes theorm gives us prob of him reporting a 6 ( false or truth) as the it caters for both scenarios

but the Q asks for reported six and truth ??

so should it be 1/8 then ???

adeel wrote:hi

the way maihuna has calculated the porb using bayes theorm gives us prob of him reporting a 6 ( false or truth) as the it caters for both scenarios

but the Q asks for reported six and truth ??

so should it be 1/8 then ???

It goes like this compute the total probability and then divide it using the component you are interested, so we found the prob of all cases and divide it with the case when it is really a six, hope i am clear, ans as 3/8 is correct as per source.

cheers

understood!!