pradhan wrote:
A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually six is
1. 1/8
2. 2/8
3. 3/8
4. 1/2
Ans is 3/8 I am unable to understand how.
Responding to a pm:
Quote:
This is how I approached the problem:
Probability of speaking truth is 3/4. And, the probability that a thrown die reports 6 is 1/6. So, the probability that it is actually six is (3/4)*(1/6) = 1/8.
As discussed above, the question is not very clear. We are not sure how he lies. If we have to make an assumption, we might assume that every time he lies, he calls it a 6 or we may assume that he lies consistently i.e. he will use every number equally while lying. So say if he throws a 1 twenty times, he will lie 5 times. When he lies, he will call it a 2 once, a 3 once, a 4 once, a 5 once and a 6 once.
Assuming he lies consistently:So consider a case when he throws the die 24 times. He gets a 1 four times, 2 four times, 3 four times and so on since all are equally likely.
Of the 4 times he throws a 1, he lies once. Say he calls it a 2.
Of the 4 times he throws a 2, he lies once. Say he calls it a 3.
Of the 4 times he throws a 3, he lies once. Say he calls it a 4.
Of the 4 times he throws a 4, he lies once. Say he calls it a 5.
Of the 4 times he throws a 5, he lies once. Say he calls it a 6.
Of the 4 times he throws a 6, he lies once. Say he calls it a 1. He says the truth in other 3 instances and calls it a 6.
We have to find the probability that when he says it is 6, it actually is a 6.
So 4 times he says it is a 6. Out of these 4 times, it is actually 6 in 3 instances.
So the probability that it actually is 6 (when he says it is 6) is 3/4.
Assuming he calls it a 6 every time he liesOf the 4 times he throws a 1, he lies once. Say he calls it a 6.
Of the 4 times he throws a 2, he lies once. Say he calls it a 6.
Of the 4 times he throws a 3, he lies once. Say he calls it a 6.
Of the 4 times he throws a 4, he lies once. Say he calls it a 6.
Of the 4 times he throws a 5, he lies once. Say he calls it a 6.
Of the 4 times he throws a 6, he lies once. Say he calls it something else. He says the truth in other 3 instances and calls it a 6.
So 8 times he says that it is a 6. Out of these 8 times, it is actually 6 in 3 instances.
So the probability that it actually is 6 (when he says it is 6) is 3/8.
You can calculate this as given by Bunuel above.
Where did you go wrong:You are finding the probability that he throws a 6 AND then calls it a 6 (P(Throws a 6) - 1/6; P(Calls it 6) - 3/4; Does both = 1/6 * 3/4 = 1/8).
But that it not the question. You have to find the probability that it is actually 6 (after he has said that it is a 6). So we need to find
P(No of time it is a 6 when he calls it a 6)/(Total number of time he calls it a 6)
It is already given that he calls it a 6. Now we have to figure the probability that it actually is a 6.
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) and though the mathematics on this question is right logically the answer still feels should be 3/4 (though it isn't an option here). Applying probability....
