Three shots are fired at a target in succession. The probabilities of a hit in the first shot is , in the second and in the third shot is , In case of exactly one hit, the probability of destroying the target is and in the case of exactly two hits an in the case of three hits is 1.0. Find the probability of destroying the target in three shots

Engineering

Mathematics

Conditional Probability

Question

Three shots are fired at a target in succession. The probabilities of a hit in the first shot is $\frac{1}{2}$ , in the second $\frac{2}{3}$ and in the third shot is $\frac{3}{4}$ , In case of exactly one hit, the probability of destroying the target is $\frac{1}{3}$ and in the case of exactly two hits $\frac{7}{11}$ an in the case of three hits is 1.0. Find the probability of destroying the target in three shots

$\frac{3}{4}$

$\frac{3}{8}$

$\frac{5}{8}$

$\frac{4}{5}$

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A : Target hit in 1st shot

B : Target hit in 2nd shot

C : Target hit in 3rd shot

E₁: destroyed in exactly one shot

E₂: destroyed in exactly two shot

E₃: destroyed in exactly three shot

$P (E_{1}) = P (E_{1} A \bar{B} \bar{C} \cup E_{1} \bar{A} \bar{B} C \cup E_{1} \bar{A} B \bar{C})$

$= \frac{1}{3} [\frac{1}{2} . \frac{1}{3} . \frac{1}{4} + \frac{1}{2} . \frac{1}{3} . \frac{3}{4} + \frac{1}{2} . \frac{2}{3} . \frac{1}{4}] = \frac{1 + 3 + 2}{3.24} = \frac{1}{12}$

$P (E_{2}) = P (E_{2} \bar{A} BC \cup E_{2} AB \bar{C} \cup E_{3} A \bar{B} C)$

$= \frac{7}{11} [\frac{1}{2} . \frac{2}{3} . \frac{3}{4} + \frac{1}{2} . \frac{2}{3} . \frac{1}{4} + \frac{1}{2} . \frac{1}{3} . \frac{3}{4}] = \frac{7.11}{11.24} = \frac{7}{24}$

$P (E_{3}) = P (E_{3} ABC) = 1 . \frac{1}{2} . \frac{2}{3} . \frac{3}{4} = \frac{1}{4}$

$P(E 1 UE 2 UE 3) = P(E 1) + P(E 2) + P(E 3)$

$= \frac{1}{12} + \frac{7}{24} + \frac{1}{4} = \frac{2 + 7 + 6}{24} = \frac{15}{24} = \frac{5}{8}$

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