The probabilities that a man makes a certain dangerous journey by car, motorcycle or by walking are 12, 16, and 13 respectively. The probability of the occurrence of an accident when he use thesemeans of transport are 15,25 and 110 respectively COLUMN-I COLUMN-III)The probability of an accident occurring in a single journey isP)16II)If an accident is known to have happened the probability that the man wastravelling by car isQ) 13 III)If an accident is known to have happened theprobability that the man was travelling by motor cycle isR) 12IV)If an accident is known to havehappened the probability that the manwas travelling on foot isS) 15

P(c)=12 P(M)=16 p(F)=13 A = accident occurs P(AB)=15 P(AM)=25 p(AF)=110P(A)=P(A∩C)+P(A∩M)+P(A∩F)= P(c). (P A/c)+P(M).P(A/M)+P(F).P(A/F)= 12.15+16.25+13.110=15P (MA)=16.2515=13P (CA)=12.1515=12P(FA)=18. 1101/5=16

The probabilities that a man makes a certain dangerous journey by car, motorcycle or by walking are 12, 16, and 13 respectively. The probability of the occurrence of an accident when he use thesemeans of transport are 15,25 and 110 respectively COLUMN-I COLUMN-III)The probability of an accident occurring in a single journey isP)16II)If an accident is known to have happened the probability that the man wastravelling by car isQ) 13 III)If an accident is known to have happened theprobability that the man was travelling by motor cycle isR) 12IV)If an accident is known to havehappened the probability that the manwas travelling on foot isS) 15

P(c)=12 P(M)=16 p(F)=13 A = accident occurs P(AB)=15 P(AM)=25 p(AF)=110P(A)=P(A∩C)+P(A∩M)+P(A∩F)= P(c). (P A/c)+P(M).P(A/M)+P(F).P(A/F)= 12.15+16.25+13.110=15P (MA)=16.2515=13P (CA)=12.1515=12P(FA)=18. 1101/5=16

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The probabilities that a man makes a certain dangerous journey by car, motorcycle or by walking
are $\frac{1}{2}$ , $\frac{1}{6}$ , and $\frac{1}{3}$ respectively. The probability of the occurrence of an accident when he use these
means of transport are $\frac{1}{5}, \frac{2}{5} and \frac{1}{10}$ respectively

COLUMN-I

COLUMN-II
I) The probability of an accident occurring in a single journey is P) $\frac{1}{6}$
II)
If an accident is known to have happened the
probability that the man was
travelling by car is
Q)
$\frac{1}{3}$

III)
If an accident is known to have happened the
probability that the man was travelling
by motor cycle is
R)

$\frac{1}{2}$
IV)
If an accident is known to have
happened the probability that the man
was travelling on foot is
S)

$\frac{1}{5}$

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$A - S, B - R, C - Q, D - P$

$A - S, B - R, C - P, D - Q$

$A - R, B - S, C - P, D - Q$

$A - Q, B - S, C - P, D - P$

answer is D.

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Detailed Solution

$P (c) = \frac{1}{2} P (M) = \frac{1}{6} p (F) = \frac{1}{3} A = a c c i d e n t o c c u r s \begin{array}{l} P (\frac{A}{B}) = \frac{1}{5} P (\frac{A}{M}) = \frac{2}{5} p (\frac{A}{F}) = \frac{1}{10} \\ \begin{array}{l} P (A) = P (A \cap C) + P (A \cap M) + P (A \cap F) \\ = P (c) . (P A / c) + P (M) . P (A / M) + P (F) . P (A / F) \\ = \frac{1}{2} . \frac{1}{5} + \frac{1}{6} . \frac{2}{5} + \frac{1}{3} . \frac{1}{10} \\ = \frac{1}{5} \\ P (\frac{M}{A}) = \frac{\frac{1}{6} . \frac{2}{5}}{\frac{1}{5}} = \frac{1}{3} \\ P (\frac{C}{A}) = \frac{\frac{1}{2} . \frac{1}{5}}{\frac{1}{5}} = \frac{1}{2} \\ P (\frac{F}{A}) = \frac{\frac{1}{8} . \frac{1}{10}}{1 / 5} = \frac{1}{6} \end{array} \end{array}$

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