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. a man speaks 3 out of 4 times.he throws a dice n report it to be 6 what is probability of being 6
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- man speaks truth 3 out of 4 times=3/4
probability of getting 6 on throwing a dice is 1/6
so reqd prob=(3/4)*(1/6)=1/8 - 11 years agoHelpfull: Yes(29) No(5)
- it should be speak truth 3 out of 4.
prob. of true speak=3/4,prob.of flase speak=1/4
prob. of getting 6=1/6
prob. of not getting 6=5/6
so prob=(3/4*1/6)/3/4*1/6+5/6*1/6
ans=3/8
- 11 years agoHelpfull: Yes(9) No(7)
- GUYZ WHICH ONE 3/8
OR 1/8
M CONFUSED????????????? - 11 years agoHelpfull: Yes(5) No(1)
- man speaks truth 3 out of 4 times, probability=3/4
probability of man speaking false = 1-3/4 =1/4
probability of getting 6 on dice=1/6
probability of not getting 6 on a dice= 1-1/6 = 5/6
therefore reqd probability= (3/4)(1/6)/ (3/4)(1/6)+(1/4)(5/6)
= 3/8
- 11 years agoHelpfull: Yes(4) No(1)
- probability of being six is= 3/4*1/6
=1/8
- 11 years agoHelpfull: Yes(3) No(2)
- probability of getting a die is=1/6
and the speaks 3/4 times so
probablity of being six=(1/6)*(3/4)=1/8 - 11 years agoHelpfull: Yes(2) No(2)
it should be speak truth 3 out of 4.
prob. of true speak=3/4,prob.of flase speak=1/4
prob. of getting 6=1/6
prob. of not getting 6=5/6
so prob=(3/4*1/6)/3/4*1/6+5/6*1/6
ans=3/8
- 11 years agoHelpfull: Yes(2) No(3)
- options-A) 3/4
B) 5/8
C) 2/5
D) 3/5
E) 4/5
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
Answer A.
- 10 years agoHelpfull: Yes(2) No(0)
- 3/4*1/6=1/8
- 11 years agoHelpfull: Yes(1) No(2)
- probability of getting 6 =1/6
probability of a speaking truth =3/4
so probability that a got 6 is= (3/4)*(1/6)= 1/8 - 11 years agoHelpfull: Yes(1) No(2)
- they asked only the probability of being 6.so i guess ans is 1/8
- 11 years agoHelpfull: Yes(1) No(1)
- If 6 actually appeared, he can report it with the probability of 3/4. If 6 has not appeared, still he can report it wrongly with the probability of 1/4
So the probability that it is actually a 6 = (Probability to appear 6 x His truthfulness to report + Probability to appear any other number x His lieing probability ) = (1/6×3/4)+(5/6×1/4)=1/3
The probability that it is actually 6 = (Probability that he reports 6)/(Total probability to appear 6)=(3/4×1/6)/((3/4×1/6)+(1/4×5/6))=3/8 - 11 years agoHelpfull: Yes(0) No(0)
- 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
- 9 years agoHelpfull: Yes(0) No(0)
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