Truth or Dare!

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gmatrant: Master | Next Rank: 500 Posts; Posts: 416; Joined: Wed Oct 03, 2007 9:08 amFollowed by: 1 member; Thanked: 10 times

by gmatrant » Fri Nov 11, 2011 12:36 pm

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

A kudos or thanks would do great if my answer has helped you

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Source: Beat The GMAT — Problem Solving | Fri Nov 11, 2011 12:36 pm

harshsheth: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sat Apr 09, 2011 10:59 am

by harshsheth » Fri Nov 11, 2011 1:15 pm

This is a case of multiple probability

Probability of him saying the truth is : 3/4
Probability of him gettin a 6 on roll : 1/6

Multiplying
3/4 * 1/6

= 1/8

Hence the correct answer is 1/8

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user123321: Master | Next Rank: 500 Posts; Posts: 385; Joined: Fri Sep 23, 2011 9:02 pmFollowed by: 6 members; Thanked: 62 times

by user123321 » Fri Nov 11, 2011 1:22 pm

IMO 3/8

The thing is if it is not six he can lie that it is six or if it is six then he can tell truth

if it is not six and he told lie => 5/6*1/4 = A(say)
if it is six and he told truth => 1/6*3/4 = B(say)

probability that it is six and he told truth is (1/6*3/4)/((1/6*3/4)+(5/6*1/4)) just like A/(A+B)
=> 3/8

user123321

Just started my preparation

Want to do it right the first time.

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gmatrant: Master | Next Rank: 500 Posts; Posts: 416; Joined: Wed Oct 03, 2007 9:08 amFollowed by: 1 member; Thanked: 10 times

by gmatrant » Fri Nov 11, 2011 3:12 pm

user123321,

if it is not six and he told lie => 5/6*1/4 = A(say)

The question asks for the probability that it is a six- "He throws a die and reports that it is a six. The probability that it is actually six is " .

So Case A is not valid isn't it, why do we need to worry if he lied.

A kudos or thanks would do great if my answer has helped you

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ArunangsuSahu: Master | Next Rank: 500 Posts; Posts: 382; Joined: Thu Mar 31, 2011 5:47 pm; Thanked: 15 times

by ArunangsuSahu » Fri Nov 11, 2011 8:59 pm

Step1: Probability of getting a six in the throw of a die=1/6
Step2: Truth = 3/4

So the combined probability = 3/4*1/6=1/8

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Fri Nov 11, 2011 9:00 pm

I agree with user123321, we need to consider both cases as the man has already reported the number as 6. But we are interested only in knowing if he said the truth.

P(getting 6/ reporting 6) = P(getting 6 and reporting six)/ [P(getting 6 and reporting six) + P(getting any other number expect 6 and reporting 6)]

= (1/6 * 3/4) /[(1/6 * 3/4) + (5/6 * 1/4)] = 3/8

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