Bob speaks truth 3 out of 4 times, He throws a die and reports that it

Question

GMATBusters’ Quant Quiz 2.0 Question -4 
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Bob speaks truth 3 out of 4 times, He throws a die and reports that it is a four. Find the probability that it is actually a four.
A. 2/8
B. 3/8
C. 4/8
D. 5/8
E. 6/8

numb007 · Accepted Answer

Ans: B
Probability that four is reported= P(He got a four and reported correctly)+P(He got another number and reported false)
P(Reporting a 4)=1/6*3/4+5/6*1/4=8/24
We have to find Probability that he actually gets a 4= Pobability that he gets an 4 / All Probability that he gets a 4
 = (1/6*3/4)/(8/24) =3/8

rvarora · Answer

Answer to this question is B - 3/8

Solution is attached herewith

pk123 · Answer

Bob speaks truth 3 out of 4 times, He throws a die and reports that it is a four. Find the probability that it is actually a four.
A. 2/8
B. 3/8
C. 4/8
D. 5/8
E. 6/8

Bob speaking truth =P(True)= 3/4
Bob speaking lie =P(Lie)= 1/4

A die landing up as 4=1/6
ie. Probability that outcome is 4 =1/6
Probability that outcome is not 4 = 5/6

Overall sample space= probability that Bob speaks truth & outcome is 4 + probability that Bob speaks lie & outcome is not 4
 =3/4*1/6 + 1/4*5/6 =8/24

Probability that bob speaks truth & outcome is 4= 3/4*1/6=3/24

Probability that it is actual 4 = Probability that bob speaks truth & outcome is 4/ (probability that Bob speaks truth & outcome is 4 + probability that Bob speaks lie & outcome is not 4)= . (3/24)/(8/24)=3/8

Option B

willacethis · Answer

probability that man speaks truth = 3/4
probability that man lies = 1/4
probability that 4 occurred = 1/6
probability that 4 did not occur = 5/6

according to Bayes' theorem

answer =  /{ + }
=3/8 = option E

p.s we can also include a case in which probability that the man randomly chooses a number to report when he lies can be included (i.e. 1/5). in that case, the denominator shall be { + } and answer would be 3/4

globaldesi · Answer

total possible outcomes of the throw
A --- when he told truth and it was a 4:  \frac{3}{4} * \frac{1}{6}  =  \frac{1}{8} 
B ---- when he lied and it was not a 4 :  \frac{1}{4} * \frac{5}{6} = \frac{5}{24} 
thus total probable cases =  \frac{1}{8 }+ \frac{5}{24}  =  \frac{1}{3}

probability when it was an actual 4 and he told the truth =  \frac{1}{8}/\frac{1}{3}  =  \frac{3}{8} 
thus B

fauji · Answer

Probability that Bob speaks Truth : 3/4

3/4 or 6/8

IMO Option E!

QuantMadeEasy · Answer

probability of getting a four is 1/6 and speaking truth is 3/4
required probability = 1/6 x 3/4 = 1/8

SUNNYRHODE002 · Answer

Answer :B

Solution in the picture attached

vishumangal · Answer

Ans is B, I just made a educated guess.

Archit3110 · Answer

truth = 3/4 
false = 1/4
reports 4 ; 1/6 and not 4 = 5/6
so Find the probability that it is actually a four = 5/6 * 3/4 ; 5/8 ; 1- 5/8 ; 3/8
IMO B 
P of getting 4 ;1/6
And if it's true then if it's actually 4 and Bob speaks truth then 3/4 considering that the truth is valid per roll of dice not total cases.

Bob speaks truth 3 out of 4 times, He throws a die and reports that it is a four. Find the probability that it is actually a four.
A. 2/8
B. 3/8
C. 4/8
D. 5/8
E. 6/8

Aviral1995 · Answer

Probability that it is not a 4 = (3/4)
Probability that it is actually 4 is 1-(3/4)=1/4 or 2/8
option A

Raxit85 · Answer

Let's annotate, Truth = T and the no. 4 is coming when bob throws dice as D (4).
So, probability of coming 4 = 1/6
So, P (T) = probability of saying truth, i.e. 3/4,
Not P (T) = probability of not saying truth, i.e. Total - saying truth, 1-3/4=1/4,

Occurrence of 4 while throwing a dice and reporting truth/false statement,

Case i) actually occurring 4 and saying truth, D (4) * P (T) = 1/6 * 3/4 = 3/24
Case ii) actually coming another no. (except 4) and reporting false as a 4
Probability of coming another no. (except 4) = 5/6
Probability of reporting false as a 4 = 1/4,
So, in this case, probability of actually coming another no. (except 4) and reporting false as a 4 = 5/6 * 1/4 = 5/24,

So, total probability of reporting 4 = case i + case ii = 3/24 + 5/24 = 8/24 = 1/3-----(1)

Now, the desired one is probability that it is actually a four, when actual 4 came and he reported it as 4, = 1/6 * 3/4 = 1/8-----(2)

Probability that it is actually a four = Desired/total = Value of eqa. (2)/Value of eqa. (1) = 1/8 divided by 1/3 = 3/8.

Hence, Ans. is B.

lacktutor · Answer

Bob speaks truth 3 out of 4 times, He throws a die and reports that it is a four. Find the probability that it is actually a four.

The Probability in which it was actually a 4 =  \frac{1}{6} 
The Probability in which he said the truth =  \frac{3}{4} 
--> The Probability that he said the truth when it was actually a 4
=  \frac{1}{6} * \frac{3}{4} =\frac{1}{8} .

Let's talk on the other possible scenario :

The Probability that it was not a 4 =  \frac{5}{6} 
Probability that he lied =  1- \frac{3}{4} = \frac{1}{4} 
--> There is something we have to mention: 
If he randomly chooses a number to report when he lies, then the probability he chooses 4 is  \frac{1}{5} .(one of 1,2,3,5 or 6)

The probability of this scenario =  \frac{5}{6} * \frac{1}{4} *\frac{1}{5} = \frac{1}{24}

well, the final probability =(1/8)/(1/8+1/24) =  \frac{3}{4}  or  \frac{6}{8}

The answer is E

hiranmay · Answer

Bob speaks truth 3 out of 4 times, He throws a die and reports that it is a four. Find the probability that it is actually a four.
A. 2/8
B. 3/8 -->  correct 
C. 4/8
D. 5/8
E. 6/8

Solution :
probability of telling the truth = P(T) = 3/4
probability of telling a lie = P(L) = 1/4
probability of getting a four = P(4) = 1/6
probability of not getting a four = P(!4) = 5/6
=> ratio or probability that Bob reports 4 = P(4)*P(T) + P(!4)*P(L)= (1/6)*(3/4) + (5/6)*(1/4)=8/24=1/3 ---(i)
=>ratio or probability that Bob tells the truth about 4 = P(4)*P(T) = (1/6)*(3/4) =1/8 ---(ii)

So the probability that it is actually a four = (ii)/(i) = (1/8)/(1/3)=3/8

sybrenblokvoort · Answer

Truth = 3/4 (T), thus Lie = 1/4 (L)

Possibilities:

Truth (3/4) and Four (1/6) = 3/4 * 1/6 = 3/24
Lie (1/4) and every number (6/6) = 1/4 * 6/6 = 6/24 +
 9/24 = 3/8

AndreV · Answer

Let's define the following events:
T= Bob reports the number obtained is 4
N= The number obtained when rolled the dice was 4

We need to find the following: P(N|T)

By Bayesian theorem,

P(N|T) =  /P(T)

Analyzing:

T/N means that, in fact, Bob is telling the truth since the number obtained was actually 4. Therefore, P(T/N)=3/4

To calculate P(T), we need the following analysis:
- if the number obtained is 4, then Bob is telling the truth. P(T)1= (probability of obtaining a 4)*(probability of telling the truth) = (1/6)*(3/4)=3/24
- if the number obtained is not 4, then Bob is lying. P(T)2= (probability of obtaining a # different from 4)*(probability of not telling the truth) = (5/6)*(1/4)=5/24

Then,

P(T)=P(T)1+P(T)2=3/24+5/24=8/24

Therefore,

P(N|T)=(1/6*3/4)/(8/24)=(3/24)/(8/24)=3/8

KarishmaB · Answer

Responding to pm:

This is the correct answer. Every time he lies, we cannot assume that he says 4. The probability of that happening is 1/5.
So answer will be 3/4.

Aviral1995 · Answer

VeritasKarishma  @BUNUEL...COULD YOU PLEASE ELABORATE MORE ON HOW YOU GOT 1/5

KarishmaB · Answer

Say, actually the die showed 1. Bob lied so he would say 2 or 3 or 4 or 5 or 6. The probability that he said 4 is 1/5 (assuming probability of any number is the same).

Bob speaks truth 3 out of 4 times, He throws a die and reports that it

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Bob speaks truth 3 out of 4 times, He throws a die and reports that it

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