Name: Probability of Random Variables
Uploaded: 2019-07-19
Duration: 10 min 25 s
Description: In this video, we will learn the formula and the method of finding the probability of random experiments with examples in detail. View the video now.

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 exactly one hit, the probability of destroying the target is

\frac{1}{3}

and in the case of exactly two hits

\frac{7}{11}

and in the case of three hits is

1.0

. Find the probability of destroying the target in three shots.

Accepted Answer

Let

$A :$ Event that the target hit in $1^{st}$ shot

$B :$ Event that the target hit in $2^{nd}$ shot

$1.0$ Event that the target hit in $3^{rd}$ shot.

$E_{1} :$ Event that the destroyed in exactly one shot

$E_{2} :$ Event that the destroyed in exactly two shot

$E_{3} :$ Event that the destroyed in exactly three shot

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

$\Rightarrow P (E_{1}) = \frac{1}{3} [\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4}]$

$\Rightarrow P (E_{1}) = \frac{1 + 3 + 2}{3^{2} \cdot 2 \cdot 4} = \frac{1}{12}$

And,

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

$\Rightarrow P (E_{2}) = \frac{7}{11} [\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4}]$

$\Rightarrow P (E_{2}) = \frac{7}{11} \cdot \frac{11}{24} = \frac{7}{24}$

And,

$P (E_{3}) = P (A B C) = 1 \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}$

Now,

$P (E_{1} \cup E_{2} \cup E_{3}) = P (E_{1}) + P (E_{2}) + P (E_{3})$

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

Important Questions on Probability

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