Question

Three persons A, B and C shoot to hit a target. If A hits the target four times in five
trials, B hits it three times in four trials and C hits it two times in three trials, find
the probability that:
$
  {\text{i}}{\text{.Exactly two persons hit the target}} \\
  {\text{ii}}{\text{.At least two persons hit the target }} \\
  {\text{iii}}{\text{. None hit the target}} \\
 $

Answer 1

Hint: In this type of problem we firstly find the respective events probabilities of hitting(${\text{P(E)}}$ or not hitting (${\text{P(}}\overline {\text{E}} ) = 1 - {\text{P(E)}}$). And then use the concept independent probability to get the desired results
According to independent probability, if there are three independent events A, B, C then probability of occurring all three id given by
$ \Rightarrow {\text{ P(A}} \cap {\text{B}} \cap {\text{C) = P(A)}}{\text{.P(B)}}{\text{.P(C)}}$
Since, all events of shooting are independent so we can use the concept of independent probability.

Complete step-by-step answer:
Since, A hits the target four times out of five trails. Let P(A) be the probability of hitting the target by person A and ${\text{P}}(\overline {\text{A}} )$be the probability that shoots do not hit the target. $ \Rightarrow {\text{P(A) = }}\dfrac{4}{5}{\text{ ----eq}}{\text{.1}}$
And
$
   \Rightarrow {\text{P(}}\overline {\text{A}} ) = 1 - {\text{P(A)}} \\
   \Rightarrow {\text{P(}}\overline {\text{A}} ) = 1 - \dfrac{4}{5} \\
   \Rightarrow {\text{P(}}\overline {\text{A}} ) = {\text{ }}\dfrac{1}{5}{\text{ -----eq}}{\text{.2}} \\
$
 Now, B hits the target three times out of four trails. Let P(B) be the probability of hitting the target by person B and ${\text{P}}(\overline {\text{B}} )$be the probability that shoots do not hit the target. $ \Rightarrow {\text{P(B) = }}\dfrac{3}{4}{\text{ -----eq}}{\text{.3}}$
And
$
   \Rightarrow {\text{P(}}\overline {\text{B}} ) = 1 - {\text{P(B)}} \\
   \Rightarrow {\text{P(}}\overline {\text{B}} ) = 1 - \dfrac{3}{4} \\
   \Rightarrow {\text{P(}}\overline {\text{B}} ) = {\text{ }}\dfrac{1}{4}{\text{ ------eq}}{\text{.4}} \\
$
Finally, C hits the target two times out of three trails. Let P(C) be the probability of hitting the target by person C and ${\text{P}}(\overline {\text{C}} )$be the probability that shoots do not hit the target. $ \Rightarrow {\text{P(C) = }}\dfrac{2}{3}{\text{ -----eq}}{\text{.5}}$
And
$
   \Rightarrow {\text{P(}}\overline {\text{C}} ) = 1 - {\text{P(C)}} \\
   \Rightarrow {\text{P(}}\overline {\text{C}} ) = 1 - \dfrac{2}{3} \\
   \Rightarrow {\text{P(}}\overline {\text{C}} ) = {\text{ }}\dfrac{1}{3}{\text{ -----eq}}{\text{.6}} \\
$
Now take the sub-problems
${\text{i}}{\text{.Exactly two persons hit the target}}$
Let ${\text{P(}}{{\text{E}}_1})$ be the probability that exactly two persons hit the target.
Then,

$ \Rightarrow {\text{P(}}{{\text{E}}_1}{\text{) = P(A}} \cap {\text{B}} \cap \overline {\text{C}} {\text{) + P(A}} \cap \overline {\text{B}} \cap {\text{C) + P(}}\overline {\text{A}} \cap {\text{B}} \cap {\text{C}}) \\
$
$
   \Rightarrow {\text{P(}}{{\text{E}}_1}{\text{) = P(A)}}{\text{.P(B}}).{\text{P(}}\overline {\text{C}} {\text{) }} + {\text{P(A}}).{\text{P(}}\overline {\text{B}} ).{\text{P(C) + P(}}\overline {\text{A}} ).{\text{P(B)}}{\text{.P(C}}){\text{ \{ }}\therefore {\text{from eq}}{\text{.1\} }} \\
$$
  {\text{Using all above formed equations we can rewrite above equation as}} \\
   \Rightarrow {\text{P(}}{{\text{E}}_1}{\text{) = }}\dfrac{4}{5}{\text{.}}\dfrac{3}{4}{\text{.}}\dfrac{1}{3}{\text{ }} + \dfrac{4}{5}.\dfrac{1}{4}{\text{.}}\dfrac{2}{3}{\text{ + }}\dfrac{1}{5}.\dfrac{3}{4}{\text{.}}\dfrac{2}{3}{\text{ }} \\
   \Rightarrow {\text{P(}}{{\text{E}}_1}{\text{) = }}\dfrac{{26}}{{60}} \\
   \Rightarrow {\text{P(}}{{\text{E}}_1}{\text{) = }}\dfrac{{13}}{{30}} \\
$
Hence, the probability that exactly two persons hit the target is $\dfrac{{13}}{{30}}$.
${\text{ii}}{\text{.At least two persons hit the target }}$
Let ${\text{P(}}{{\text{E}}_2})$ be the probability that at least two persons hit the target.
Then,
$ \Rightarrow {\text{P(}}{{\text{E}}_2}{\text{) = P(A}} \cap {\text{B}} \cap \overline {\text{C}} {\text{) + P(A}} \cap \overline {\text{B}} \cap {\text{C) + P(}}\overline {\text{A}} \cap {\text{B}} \cap {\text{C}}) + {\text{P(A}} \cap {\text{B}} \cap {\text{C)}} \\
   \Rightarrow {\text{P(}}{{\text{E}}_2}{\text{) = P(A)}}{\text{.P(B}}).{\text{P(}}\overline {\text{C}} {\text{) }} + {\text{P(A}}).{\text{P(}}\overline {\text{B}} ).{\text{P(C) + P(}}\overline {\text{A}} ).{\text{P(B)}}{\text{.P(C}}){\text{ + P(A)}}{\text{.P(B)}}{\text{.P(C) \{ }}\therefore {\text{from eq}}{\text{.1\} }} \\
  {\text{Using all above formed equations we can rewrite above equation as}} \\
   \Rightarrow {\text{P(}}{{\text{E}}_2}{\text{) = }}\dfrac{4}{5}{\text{.}}\dfrac{3}{4}{\text{.}}\dfrac{1}{3}{\text{ }} + \dfrac{4}{5}.\dfrac{1}{4}{\text{.}}\dfrac{2}{3}{\text{ + }}\dfrac{1}{5}.\dfrac{3}{4}{\text{.}}\dfrac{2}{3}{\text{ + }}\dfrac{4}{5}{\text{.}}\dfrac{3}{4}{\text{.}}\dfrac{2}{3}{\text{ }} \\
   \Rightarrow {\text{P(}}{{\text{E}}_2}{\text{) = }}\dfrac{{50}}{{60}} \\
   \Rightarrow {\text{P(}}{{\text{E}}_2}{\text{) = }}\dfrac{5}{6} \\
$
Hence, the probability that at least two persons hit the target is $\dfrac{5}{6}$.
${\text{iii}}{\text{. None hit the target}}$
Let ${\text{P(}}{{\text{E}}_3})$ be the probability that none of persons hit the target.
Then,
$
   \Rightarrow {\text{P(}}{{\text{E}}_3}){\text{ }} = {\text{ P(}}\overline {\text{A}} \cap \overline {\text{B}} \cap \overline {\text{C}} ) \\
   \Rightarrow {\text{P(}}{{\text{E}}_3}){\text{ }} = {\text{ P(}}\overline {\text{A}} ).{\text{P(}}\overline {\text{B}} ).{\text{P(}}\overline {\text{C}} ) \\
   \Rightarrow {\text{P(}}{{\text{E}}_3}){\text{ }} = {\text{ }}\dfrac{1}{5}.\dfrac{1}{4}.\dfrac{1}{3} \\
   \Rightarrow {\text{P(}}{{\text{E}}_3}){\text{ }} = {\text{ }}\dfrac{1}{{60}} \\
$
Hence, the probability that none of persons hit the target is $\dfrac{1}{{60}}$.
Note:- Whenever you get this type of problem the key concept to solve this is to have knowledge about independent probability and its implementations. And one more thing to be noted is that independent probability does not depend on occurrence of any other event.