Probability-Total Probability & Bayes Theorem(15 Jan) - Free MCQ Practice

Let E be the event that the man reports that six occurs in the throwing of the die and 
let S1 be the event that six occurs and S2 be the event that six does not occur.  Then P(S1) = Probability that six occurs = 1/6 
P(S2) = Probability that six does not occur = 5/6 
P(E|S1) = Probability that the man reports that six occurs when six has actually occurred on the die 
= Probability that the man speaks the truth = 3/4 
P(E|S2) = Probability that the man reports that six occurs when six has not actually occurred on the die 
= Probability that the man does not speak the truth = 1 - 3/4 = 1/4 
Thus, by Bayes' theorem, we get 
 P(S1|E) = Probability that the report of the man that six has occurred is actually a six 
⇒P(S1/E) = (P(S1)P(E/S1))/(P(S1)P(E/S1) + P(S2)P(E/S2)) 
= (1/6 x 3/4)/(1/6 x 3/4 + 5/6 x 1/4) 
= 1/8 x 24/8 
= 3/8

 Let E : A speaks truth
F : A lies
H : head appears on the toss of coin
P(E) = Probability A speaks truth 
= 5/9
P(F) = Probability A speaks lies
1 - P(E) => 1-(5/9) 
=> 4/9
P(H/E) = Probability that appears head, if A speaks truth  = ½
P(E/H) = Probability that appears head, if A speaks lies  = 1/2
P(H/E) = [(5/9) (½)]/[(5/9) (½) + (4/9) (½)]
= (5/18)/[(5/18) + (4/18)]
= 5/9
 Let E : A speaks truth
F : A lies
H : head appears on the toss of coin
P(E) = Probability A speaks truth 
= 5/9
P(F) = Probability A speaks lies
1 - P(E) => 1-(5/9) 
=> 4/9
P(H/E) = Probability that appears head, if A speaks truth  = ½
P(E/H) = Probability that appears head, if A speaks lies  = 1/2
P(H/E) = [(5/9) (½)]/[(5/9) (½) + (4/9) (½)]
= (5/18)/[(5/18) + (4/18)]
= 5/9

Let E1,E2,E3andE4 be the events that the doctor comes by train, bus, scooter and car respectively. Then,
P(E1)=3/10,P(E2)=1/5,P(E3)=1/10andP(E4)=2/5.
Let E be the event that the doctor is late. Then,
P(E/E1) = probability that the doctor is late, given that he comes by train
=1/4.
P(E/E2)= probability that the doctor is late, given that he comes by bus
=1/3.
P(E/E3)= probability that the doctor is late, given that the comes by scooter
=1/12.
P(E/E4)= probability that the doctor is late, given that that he comes by car
=0.
Probability that he comes by train, given that he is late
P(E2/E)
[P(E2).P(E/E2)]/[P(E1).P(E/E1)+P(E2).P(E/E2)+P(E3).P(E/E3)+P(E4).P(E/E40)][by Bayes's theorem]
[(1/5×1/3)]/(3/10×1/4)+(1/5×1/3)+(1/10×1/12)+(2/5×0)
=(3/40×120/18)=4/9
Hence, the required probability is 4/9.

If we assume that it is equally likely that the ball is drawn from either bag, then we proceed as follows. Let R be the event that a red ball is drawn, and let B be the event that bag B is chosen for the drawing. Then
P(B) = P(not B) = 1/2
P(R|B) = 5/9 = P(RB) / P(B) = 2 P(RB),
P(R|not B) = 3/5 = P(R not B) / P(not B) = 2 P(R not B)
P(R) = P(RB) + P(R not B) = (1/2)( 5/9 + 3/5 ) = 26/45
P(B|R) = P(RB) / P(R) = (5/9)(1/2) / (26/45) = 25/52