A man speaks truth 3 out of 4 times He throws a die and reports it to be a 6

Assuming the man speaks truth always, probability for number 6 of die is clearly 1/6. Now man speaks truth 3 out of 4 times which means his probability of speaking truth is 3/4. Multiply both (1/6)(3/4) = 3/24 = 1/8
13 years agoHelpfull: Yes(24) No(37)
the probability of showing a 6 on the dice=1/6
now the man speaks truth 3/4 times
so the probability that it will be 6is (1/6)*(3/4)=1/8
13 years agoHelpfull: Yes(21) No(25)
Here n(a) is telling the truth
n(b) means false

so n(a) = 3/4 * 1/6 [3/4 prob for truth and 1/6 means possibility of 6]
so n(b) = 1/4 * 5/6 [1/4 prob for false and 5/6 means possibility of not 6]
n(s) = n(a) + n(b) = 3/4 * 1/6 + 1/4 * 5/6 = 1/3

so now prob(truth) i.e p(a) = n(a)/n(s) = (1/8)/(1/3) = 3/8 simple
9 years agoHelpfull: Yes(13) No(0)
Probability that the die does not report 6, and he is telling the truth: (5/6)*(3/4)=5/8
Therefore, Probability that the die reports 6 = 1-(5/8) = 3/8
Answer: 3/8
9 years agoHelpfull: Yes(6) No(0)
i didn't get ur ans wt is meant by" he reports as six"
9 years agoHelpfull: Yes(0) No(0)
3 out of 4 so 3/8
8 years agoHelpfull: Yes(0) No(1)
P[6 rolled & reports as 6 ] = 1/6 * 3/4 = 1/8
P[non-6 & reports as 6] = 5/6 * 1/4 = 5/24
P[reported as a 6 ] = 1/8 + 5/24 = 1/3

P[6 rolled | reported as 6] = P[6 roled & reports as 6] / P[reports as 6]
= [1/8] / [1/3] = 3/8
6 years agoHelpfull: Yes(0) No(0)

A man speaks truth 3 out of 4 times. He throws a die and reports it to be a 6.
What is the probability of it being a 6?