A man is known to speak truth 3 out of 4 times .he throws a die and a number other than six comes up .Find the probability that he reports is six
Answers
Answer:
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Step-by-step explanation:
Probability that man speaks the truth is
4
3
.
The probability that man lies is
4
1
.
Probability of getting a 6 =
6
1
Probability of not getting a six =
6
5
Applying Baye's theorem, we get the required probability as
6
1
.
4
3
+
6
5
.
4
1
6
1
.
4
3
=
8
3
.
Given : A man is known to speak truth 3 out of 4 times.
He throws a die and a number other than six comes up.
To Find : the probability that he reports it is a six
Solution:
Probability of truth = 3/4
Probability of Lie = 1 - 1/4 = 1/4
throws a die and a number other than six comes up.
Case 1 : If speak Truth then there is no probability of reporting it is a six
(3/4) (0) = 0
Case 2 If speak lie then there check probability
any number comes other than 6
so he can speak any one number other than received number
so there are 5 numbers and probability speaking 6 out of them is 1/5
Probability = (1/4) (1/5)
= 1/20
1/20 is the probability that he reports it 6
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