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

• 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
• 10 years agoHelpfull: Yes(43) No(6)
• 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
• 10 years agoHelpfull: Yes(21) No(3)
• 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
• 10 years agoHelpfull: Yes(6) No(0)
• Required probability=(3/4*1/6)/{(3/4*1/6)+(1/4*5/6)}=3/8
• 10 years agoHelpfull: Yes(5) No(3)
• 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
• 10 years agoHelpfull: Yes(3) No(1)
• 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..
• 10 years agoHelpfull: Yes(2) No(20)
• i think answer is 3/8.
• 10 years agoHelpfull: Yes(2) No(4)
• required probability=probability of speaking truth*prob of getting a 6 from dice=1/4*1/6=1/24
• 10 years agoHelpfull: Yes(1) No(9)
• ans:-3/4 as the probability of his saying truth is 3/4
• 10 years agoHelpfull: Yes(1) No(2)
• ans:1|6 since actual probability of getting 6 in thrown die is 1of 6.
• 10 years agoHelpfull: Yes(0) No(11)
• the chnc die to be 6 is 1/6
  the chnc of he spk truth s 3/4
  1/8 will be anw
• 10 years agoHelpfull: Yes(0) No(6)
• answer is 1/8
• 10 years agoHelpfull: Yes(0) No(4)