The probabilities that a man makes a certain dangerous journey by car, motor cycle or on foot are 12,16 and 13respectively. The probabilities of an accident when he uses these means of transport are 15,25 and 110respectively.

B1 = Journey by carB2 = Journey by motor cycleB3 = Journey on footPB1=12;PB2=16;PB3=13 Let A = accident occursPAB1=15;PAB2=25;PAB3=110P(A)=PA∩B1+PA∩B2+PA∩B3=PB1⋅PAB1+PB2⋅PAB2+PB3PAB3=15PB2A=16⋅2515=230⋅51=13

The probabilities that a man makes a certain dangerous journey by car, motor cycle or on foot are 12,16 and 13respectively. The probabilities of an accident when he uses these means of transport are 15,25 and 110respectively.

B1 = Journey by carB2 = Journey by motor cycleB3 = Journey on footPB1=12;PB2=16;PB3=13 Let A = accident occursPAB1=15;PAB2=25;PAB3=110P(A)=PA∩B1+PA∩B2+PA∩B3=PB1⋅PAB1+PB2⋅PAB2+PB3PAB3=15PB2A=16⋅2515=230⋅51=13

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The probabilities that a man makes a certain dangerous journey by car, motor cycle or on foot are $\frac{1}{2}, \frac{1}{6} and \frac{1}{3}$ respectively. The probabilities of an accident when he uses these means of transport are $\frac{1}{5}, \frac{2}{5} and \frac{1}{10}$ respectively.

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The probability of an accident occurring in a single journey, is

If an accident is known to have happened, the probability that the man was traveling by car, is

If an accident is known to have happened, the probability that the man was traveling by motor cycle, is

If an accident is known to have happened, the probability that the man was traveling on foot, is

answer is , , A, .

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Detailed Solution

B₁ = Journey by car
B₂ = Journey by motor cycle
B₃ = Journey on foot
$P (B_{1}) = \frac{1}{2}; P (B_{2}) = \frac{1}{6}; P (B_{3}) = \frac{1}{3}$

Let A = accident occurs
$\begin{array}{l} P (\frac{A}{B_{1}}) = \frac{1}{5}; P (\frac{A}{B_{2}}) = \frac{2}{5}; P (\frac{A}{B_{3}}) = \frac{1}{10} \\ P (A) = P (A \cap B_{1}) + P (A \cap B_{2}) + P (A \cap B_{3}) \\ = P (B_{1}) \cdot P (\frac{A}{B_{1}}) + P (B_{2}) \cdot P (\frac{A}{B_{2}}) + P (B_{3}) P (\frac{A}{B_{3}}) = \frac{1}{5} \\ P (\frac{B_{2}}{A}) = \frac{\frac{1}{6} \cdot \frac{2}{5}}{\frac{1}{5}} = \frac{2}{30} \cdot \frac{5}{1} = \frac{1}{3} \end{array}$