Latest TCS Aptitude Question SOLUTION: A man is known to speak truth 3 out of 4 times. He throws die and reports that it is a 6. The probability that it is actually a 6 is

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A man is known to speak truth 3 out of 4 times. He throws die and reports that it is a 6. The probability that it is actually a 6 is

Asked In TCS Dipannita Adhikary (11 years ago)

Solved Krati Agarwal Read Solution (12)

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Read Solution (12) :

Krati Agarwal  (11 years ago)

this is a problem in conditional probability

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

Is this solution Helpfull? Yes (43) | No (6)  

Karthika  (11 years ago)

As he has already reported that the die's a 6, either he is lying about it or it is actually a six.
Two possiblities
1.Either it is truly a six & he's telling the truth
2.It's a not a six & he's lying by saying that it's a six.

Our question is probability that it is actually a six(speaking truth)
Net Probability = (prob. of actual 6 & truth) / [prob. of actual 6 & truth + prob. of non 6 & lies ]
= (3/4)(1/6) / [3/4*1/6 + 1/4*5/6]
= 3/8

Is this solution Helpfull? Yes (21) | No (3)  

karthik  (11 years ago)

Prob that its a 6 and he speaks truth :- 3/4 * 1/6 = 3/24

Now what is the prob of reporting a 6 on a dice when it is a lie = 1/4 (speaking a lie) * 5/6 (prob of no. other than 6) * 1/5 (lieing the number as 6 and not telling any othe number except the actual no.) = 5/120 = 1/24

Thus the prob of reporting no. as 6 (total lie + truth) = 3/24 + 1/24 = 4/24 = 1/6

Thus prob of actually having 6 when reported so is = 3/24/1/6 = 3/4

Is this solution Helpfull? Yes (6) | No   

Arindam Sural  (11 years ago)

Required probability=(3/4*1/6)/{(3/4*1/6)+(1/4*5/6)}=3/8

Is this solution Helpfull? Yes (5) | No (3)  

karthik  (11 years ago)

if he is telling the truth 3 out of 4 times, that means hes telling the truth 75% of the time. That means on any give statement, he is telling the truth 75% of the time .... i.e. 0.75, which is still just 3/4

Is this solution Helpfull? Yes (3) | No (1)  

Pradeep Kumar  (11 years ago)

Prob that its a 6 and he speaks truth :- 3/4 * 1/6 = 3/24

Now what is the prob of reporting a 6 on a dice when it is a lie = 1/4 (speaking a lie) * 5/6 (prob of no. other than 6) * 1/5 (lieing the number as 6 and not telling any othe number except the actual no.) = 5/120 = 1/24

Thus the prob of reporting no. as 6 (total lie + truth) = 3/24 + 1/24 = 4/24 = 1/6

Thus prob of actually having 6 when reported so is = 3/24/1/6 = 3/4
so,answer would be 3/4..

Is this solution Helpfull? Yes (2) | No (20)  

k.m  (11 years ago)

i think answer is 3/8.

Is this solution Helpfull? Yes (2) | No (4)  

saurabh   (11 years ago)

required probability=probability of speaking truth*prob of getting a 6 from dice=1/4*1/6=1/24

Is this solution Helpfull? Yes (1) | No (9)  

vishnu  (11 years ago)

ans:-3/4 as the probability of his saying truth is 3/4

Is this solution Helpfull? Yes (1) | No (2)  

prabhurajan  (11 years ago)

ans:1|6 since actual probability of getting 6 in thrown die is 1of 6.

Is this solution Helpfull? Yes | No (11)  

dedeepya chowdary  (11 years ago)

the chnc die to be 6 is 1/6
the chnc of he spk truth s 3/4
1/8 will be anw

Is this solution Helpfull? Yes | No (6)  

nikhil pande  (11 years ago)

answer is 1/8

Is this solution Helpfull? Yes | No (4)  

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