A prisoner escapes from a jail and is equally likely to choose one of the four roads I, II, III or IV to reach away from the hands of law. If he choose I road, he is successful with probability 16 and for II, III and IV this is 18, 110 and 112. If the prisoner is successful, the probability that he chose roadColumn–IColumn–IIA)IP)1257B)IIQ)1557C)IIIR)2057D)IVS)1057

P(E1)=P(E2)=P(E3)=P(E4)=14 P(SE1)=16,P(SE2)=18,P(SE3)=110 and P(SE4)=112 P(S)=P(S∩E1)∪(S∩E2)∪(S∩E3)∪(S∩E4) =p(S∩E1)+P(S∩E2)+P(S∩E2)+P(S∩E4) =(14)(16)+(14)(18)+(14)(110)+(14)(112) =14(57120)P(E1S)=P(E1∩S)P(S)=(14)(18)(14)(57120)=2057 P(E2S)=P(E2∩S)P(S)=(14)(18)(14)(57120)=1557 P(E3S)=P(E3∩S)P(S)=(14)(110)(14)(57120)=1257 P(E4S)=P(E4∩S)P(S)=(14)(112)(14)(57120)=1057

A prisoner escapes from a jail and is equally likely to choose one of the four roads I, II, III or IV to reach away from the hands of law. If he choose I road, he is successful with probability 16 and for II, III and IV this is 18, 110 and 112. If the prisoner is successful, the probability that he chose roadColumn–IColumn–IIA)IP)1257B)IIQ)1557C)IIIR)2057D)IVS)1057

P(E1)=P(E2)=P(E3)=P(E4)=14 P(SE1)=16,P(SE2)=18,P(SE3)=110 and P(SE4)=112 P(S)=P(S∩E1)∪(S∩E2)∪(S∩E3)∪(S∩E4) =p(S∩E1)+P(S∩E2)+P(S∩E2)+P(S∩E4) =(14)(16)+(14)(18)+(14)(110)+(14)(112) =14(57120)P(E1S)=P(E1∩S)P(S)=(14)(18)(14)(57120)=2057 P(E2S)=P(E2∩S)P(S)=(14)(18)(14)(57120)=1557 P(E3S)=P(E3∩S)P(S)=(14)(110)(14)(57120)=1257 P(E4S)=P(E4∩S)P(S)=(14)(112)(14)(57120)=1057

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A prisoner escapes from a jail and is equally likely to choose one of the four roads I, II, III or IV to reach away from the hands of law. If he choose I road, he is successful with probability $\frac{1}{6}$ and for II, III and IV this is $\frac{1}{8}, \frac{1}{10}$ and $\frac{1}{12}$ . If the prisoner is successful, the probability that he chose road
Column–I
Column–II
A)
I
P)
$\frac{12}{57}$
B)
II
Q)
$\frac{15}{57}$
C)
III
R)
$\frac{20}{57}$
D)
IV
S)
$\frac{10}{57}$

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A–S, B–P, C–R, D–Q

A–Q, B–P, C–Q, D–R

A–R, B–S, C–R, D–P

A–R, B–Q, C–P, D–S

answer is A.

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Detailed Solution

$P (E_{1}) = P (E_{2}) = P (E_{3}) = P (E_{4}) = \frac{1}{4} P (\frac{S}{E_{1}}) = \frac{1}{6}, P (\frac{S}{E_{2}}) = \frac{1}{8}, P (\frac{S}{E_{3}}) = \frac{1}{10} a n d P (\frac{S}{E_{4}}) = \frac{1}{12} P (S) = P (S \cap E_{1}) \cup (S \cap E_{2}) \cup (S \cap E_{3}) \cup (S \cap E_{4}) = p (S \cap E_{1}) + P (S \cap E_{2}) + P (S \cap E_{2}) + P (S \cap E_{4}) = (\frac{1}{4}) (\frac{1}{6}) + (\frac{1}{4}) (\frac{1}{8}) + (\frac{1}{4}) (\frac{1}{10}) + (\frac{1}{4}) (\frac{1}{12}) = \frac{1}{4} (\frac{57}{120})$

$P (\frac{E_{1}}{S}) = \frac{P (E_{1} \cap S)}{P (S)} = \frac{(\frac{1}{4}) (\frac{1}{8})}{(\frac{1}{4}) (\frac{57}{120})} = \frac{20}{57} P (\frac{E_{2}}{S}) = \frac{P (E_{2} \cap S)}{P (S)} = \frac{(\frac{1}{4}) (\frac{1}{8})}{(\frac{1}{4}) (\frac{57}{120})} = \frac{15}{57} P (\frac{E_{3}}{S}) = \frac{P (E_{3} \cap S)}{P (S)} = \frac{(\frac{1}{4}) (\frac{1}{10})}{(\frac{1}{4}) (\frac{57}{120})} = \frac{12}{57} P (\frac{E_{4}}{S}) = \frac{P (E_{4} \cap S)}{P (S)} = \frac{(\frac{1}{4}) (\frac{1}{12})}{(\frac{1}{4}) (\frac{57}{120})} = \frac{10}{57}$