Question

One shot is fired from each of the three guns. Let \[A,B,C\] denote the events that the target is hit by the first, second and third gun respectively. Assuming that \[A,B,C\] are independent events and that \[P(A) = 0.5,P(B) = 0.6,P(C) = 0.8\] the probability that at least one hit is registered is?
A. \[0.96\]
B. \[0.86\]
C. \[0.76\]
D. \[0.66\]

Answer 1

Hint: Using the concept of independent events in probability i.e. the events which are not affected by previous events we solve for the value of intersection of each pair of events and value of intersection of all three events which we substitute in the formula for union of all events to get our value.
* Probability of any event is defined as the number of possibilities divided by total number of possibilities and probability of an event is always greater than or equal to zero and is always less than or equal to one.
* If three events \[A,B,C\] are given then \[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)\]
* If \[A,B,C\] are independent events then,
\[
  P(A \cap B) = P(A).P(B) \\
  P(B \cap C) = P(B).P(C) \\
  P(C \cap A) = P(C).P(A) \\
  P(A \cap B \cap C) = P(A).P(B).P(C) \\
 \]

Complete step-by-step answer:
Since we know \[A,B,C\] are the events representing target hit by first, second and third gun respectively, and \[P(A) = 0.5,P(B) = 0.6,P(C) = 0.8\] and that the events are independent events.
We can find the value of intersection of events by the formula of independent events.
\[
  P(A \cap B) = P(A).P(B) \\
  P(B \cap C) = P(B).P(C) \\
  P(C \cap A) = P(C).P(A) \\
  P(A \cap B \cap C) = P(A).P(B).P(C) \\
 \]
Here \[P(A) = 0.5,P(B) = 0.6,P(C) = 0.8\], therefore substituting the values in the equation
\[
  P(A \cap B) = P(A).P(B) \\
  P(A \cap B) = 0.5 \times 0.6 = 0.3 \\
 \]
\[
  P(B \cap C) = P(B).P(C) \\
  P(B \cap C) = 0.6 \times 0.8 = 0.48 \\
 \]
\[
  P(C \cap A) = P(C).P(A) \\
  P(C \cap A) = 0.8 \times 0.5 = 0.4 \\
 \]
\[
  P(A \cap B \cap C) = P(A).P(B).P(C) \\
  P(A \cap B \cap C) = 0.5 \times 0.6 \times 0.8 = 0.24 \\
 \]
To find the probability that at least one hit is registered we find the probability of union of all the events because at least one hit means probability of one hit, two hits and three hits all combined.
Now we substitute all the values obtained in the formula for the union
\[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)\]
                             \[
   = 0.5 + 0.6 + 0.8 - 0.3 - 0.48 - 0.4 + 0.24 \\
   = 2.14 - 1.18 \\
   = 0.96 \\
 \]
Therefore, probability that at least one shot is registered is \[0.96\]

So, the correct answer is “Option A”.

Note: Students are likely to make mistakes in the calculation part because there are decimal present, students can convert decimal into fraction if they are more comfortable with it.