Question

$A$ can hit the target three times in six shots, $B$ can hit the target 2 times in six shots and $C$ can hit the target 4 times in six shots. They fix a ball. But what is the probability that they hit the ball at least two shots.

Answer 1

Hint:This is a problem related to Probability. Here, the phrase “at least” has come, it means we have to find out that the probability of hitting the target should be greater than or equal to two. It means we shall have to find out all the events when the target is hit by 2 or more times and its probability of occurrence.

Complete step-by-step answer:
For a random event, the probability of an event, say p(E) is defined as the ratio of the number of favorable outcomes or chances and the total number of outcomes. Here,$P(E) = \dfrac{{N(A)}}{{N(S)}}$, where $N(A)$ is number of favorable outcomes and $N(S)$ is the total number of outcomes.

As we know that $A$ can hit the target three times in six shots,
Then, the probability of $A$ hitting the target is
$
   \Rightarrow P(A) = \dfrac{3}{6} \\
   \Rightarrow P(A) = \dfrac{1}{2}{\text{ }}..................{\text{ (1)}} \\
 $
And the probability of $A$ not hitting the target is
$
   \Rightarrow P(\overline A ) = 1 - \dfrac{1}{2} \\
   \Rightarrow P(\overline A ) = \dfrac{1}{2}{\text{ }}...............{\text{ (2)}} \\
 $
Similarly, the probability of $B$ hitting the target is
$
   \Rightarrow P(B) = \dfrac{2}{6} \\
   \Rightarrow P(B) = \dfrac{1}{3}{\text{ }}................{\text{ (3)}} \\
 $
As these events are complementary means they have only two possible outcomes,then we can write $P\text{(hitting the target)}$ + $P\text{(not hitting the target )}$= $1$
So, the probability of $B$ not hitting the target is
$
   \Rightarrow P(\overline B ) = 1 - \dfrac{1}{3} \\
   \Rightarrow P(\overline B ) = \dfrac{2}{3}{\text{ }}...................{\text{ (4)}} \\
 $
And, finding out the same probabilities for $C$ as the probability of $C$ hitting the target is
$
   \Rightarrow P(C) = \dfrac{4}{6} \\
   \Rightarrow P(C) = \dfrac{2}{3}{\text{ }}................{\text{ (5)}} \\
 $
And the probability of $C$ not hitting the target is
$
   \Rightarrow P(\overline C ) = 1 - \dfrac{2}{3} \\
   \Rightarrow P(\overline C ) = \dfrac{1}{3}{\text{ }}...................{\text{ (6)}} \\
 $
Now, if $A$, $B$ and $C$ altogether hit the target then the chances that at least 2 shots hit the target will be expressed by the following expression,

$P\text{(atleast two shots)}$ = $P\text{(A,B and not C)}$ + $P\text{(A,C and not B)}$ + $P\text{(B,C and not A)}$ + $P\text{(all three shots)}$.
\[ \Rightarrow [P(A) \cap P(B) \cap P(\overline C )] \cup [P(A) \cap P(\overline B ) \cap P(C)] \cup [P(\overline A ) \cap P(B) \cap P(C)] \cup [P(A) \cap P(B) \cap P(C)]\]Since $A$, $B$ and $C$ are independently hitting the target, therefore above expression will become as
\[
   \Rightarrow [P(A) \times P(B) \times P(\overline C )] + [P(A) \times P(\overline B ) \times P(C)] + [P(\overline A ) \times P(B) \times P(C)] + [P(A) \times P(B) \times P(C)] \\
   \Rightarrow P(A)P(B)P(\overline C ) + P(A)P(\overline B )P(C) + P(\overline A )P(B)P(C) + P(A)P(B)P(C) \\
 \]
Now, putting the values from equations (1) to (6), we will get
$
   \Rightarrow \dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{3} + \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{2}{3} + \dfrac{1}{2}.\dfrac{1}{3}.\dfrac{2}{3} + \dfrac{1}{2}.\dfrac{1}{3}.\dfrac{2}{3} \\
   \Rightarrow \dfrac{1}{{18}} + \dfrac{4}{{18}} + \dfrac{2}{{18}} + \dfrac{2}{{18}} \\
   \Rightarrow \dfrac{9}{{18}} = \dfrac{1}{2} \\
 $
Therefore the required probability is $\dfrac{1}{2}$.

Note:Solving these types of problems, you should take care of two phrases “at least” and “at most”. Both these have different meanings for Probability calculations. Read the number of events (in current case, no. of shots) carefully before attempting the problem.