Question

The probability that three men hit a target are \[0.3\], \[0.5\], and \[0.4\] respectively. Each man fires once at the target. (Assume that the three events that each hits the target are independent).
(a) Find the probability that they all: (i) hit the target; (ii) miss the target.
(b) Find the probability that the target is hit: (i) at least once; (ii) exactly once.
(c) If only one hits the target, what is the probability that it was the first man?

Answer 1

Hint: The probability that three men hit a target are \[0.3\], \[0.5\], and \[0.4\] respectively. Each man fires once at the target. (Assume that the three events that each hits the target are independent).
(a) Find the probability that they all: (i) hit the target; (ii) miss the target.
(b) Find the probability that the target is hit: (i) at least once; (ii) exactly once.
(c) If only one hits the target, what is the probability that it was the first man?

Complete step-by-step answer:
Let A be the event that the first man hits the target, B be the event that the second man hits the target, and C be the event that the third man hits the target.
The probability that A hits the target is \[P\left( A \right) = 0.3\].
The probability that B hits the target is \[P\left( B \right) = 0.5\].
The probability that C hits the target is \[P\left( C \right) = 0.4\].
The event that a man hits the target is complement to the event of missing the target.
Therefore, we get
The probability that A misses the target is \[P\left( {A'} \right) = 1 - P\left( A \right) = 1 - 0.3 = 0.7\].
The probability that B misses the target is \[P\left( {B'} \right) = 1 - P\left( B \right) = 1 - 0.5 = 0.5\].
The probability that C misses the target is \[P\left( {C'} \right) = 1 - P\left( C \right) = 1 - 0.4 = 0.6\].
(a)
(i) First, we will find the probability that they all hit the target.
Since the events that they each hit the target are independent, we can multiply their individual probabilities of hitting the target to get the probability that each of them hits the target.
Therefore, we get
\[P\left( {ABC} \right) = P\left( A \right)P\left( B \right)P\left( C \right)\]
Substituting \[P\left( A \right) = 0.3\], \[P\left( B \right) = 0.5\], and \[P\left( C \right) = 0.4\], we get
\[ \Rightarrow P\left( {ABC} \right) = 0.3 \times 0.5 \times 0.4\]
Multiplying the terms in the expression, we get
\[ \Rightarrow P\left( {ABC} \right) = 0.06\]
Therefore, the probability that each of them hits the target is \[0.06\].
(ii) Now, we will find the probability that they all hit the target.
Since the events that they each miss the target are independent, we can multiply their individual probabilities of missing the target to get the probability that each of them misses the target.
Therefore, we get
\[P\left( {A'B'C'} \right) = P\left( {A'} \right)P\left( {B'} \right)P\left( {C'} \right)\]
Substituting \[P\left( {A'} \right) = 0.7\], \[P\left( {B'} \right) = 0.5\], and \[P\left( {C'} \right) = 0.6\], we get
\[ \Rightarrow P\left( {A'B'C'} \right) = 0.7 \times 0.5 \times 0.6\]
Multiplying the terms in the expression, we get
\[ \Rightarrow P\left( {A'B'C'} \right) = 0.21\]
Therefore, the probability that each of them misses the target is \[0.21\].
(b)
(i) The event that the target is hit at least once is the complement of the event that the target is hit 0 times.
The event that the target is hit 0 times is the same as the event that all three men miss the target.
Therefore, we get
The probability that the target is hit at least once \[ = 1 - P\left( {A'B'C'} \right)\].
Substituting \[P\left( {A'B'C'} \right) = 0.21\], we get
The probability that the target is hit at least once \[ = 1 - 0.21\].
Subtracting the terms in the expression, we get
The probability that the target is hit at least once \[ = 0.79\].
Therefore, the probability that the target is hit at least once is \[0.79\].
(ii)
The probability that the target is hit exactly once is the sum of the probabilities of the events \[AB'C'\], \[A'BC'\], and \[A'B'C\].
Therefore, we get
The probability that the target is hit exactly once \[ = P\left( {AB'C'} \right) + P\left( {A'BC'} \right) + P\left( {A'B'C} \right)\]
Since the events are independent, we can rewrite the expression as
The probability that the target is hit exactly once \[ = P\left( A \right)P\left( {B'} \right)P\left( {C'} \right) + P\left( {A'} \right)P\left( B \right)P\left( {C'} \right) + P\left( {A'} \right)P\left( {B'} \right)P\left( C \right)\]
Substituting \[P\left( A \right) = 0.3\], \[P\left( B \right) = 0.5\], \[P\left( C \right) = 0.4\], \[P\left( {A'} \right) = 0.7\], \[P\left( {B'} \right) = 0.5\], and \[P\left( {C'} \right) = 0.6\] in the expression, we get
The probability that the target is hit exactly once \[ = 0.3 \times 0.5 \times 0.6 + 0.7 \times 0.5 \times 0.6 + 0.7 \times 0.5 \times 0.4\]
Multiplying the terms in the expression, we get
The probability that the target is hit exactly once \[ = 0.09 + 0.21 + 0.14\]
Adding the terms in the expression, we get
The probability that the target is hit exactly once \[ = 0.44\]
Therefore, the probability that the target is hit exactly once is \[0.44\].
(c)
The probability that the target is hit exactly once is \[0.44\].
The probability that the first man hits the target if the target is hit exactly once is \[P\left( {AB'C'} \right) = 0.09\].
Therefore, we can find the probability that if the target is hit, then it is hit by the first man using the expression
Required probability \[ = \dfrac{{P\left( {AB'C'} \right)}}{{P\left( {AB'C'} \right) + P\left( {A'BC'} \right) + P\left( {A'B'C} \right)}}\]
Therefore, we get
Required probability \[ = \dfrac{{0.09}}{{0.44}}\]
Simplifying the expression, we get
Required probability \[ = \dfrac{9}{{44}}\]
Therefore, if only one man hits the target, then the probability that it was the first man is \[\dfrac{9}{{44}}\].

Note: We used the complement of an event to easily calculate the probability of a man missing the target, and the probability that the target is hit at least once. The probability of an event is equal to the probability of its complement event, subtracted from 1.