The probability of hitting the target by a trained sniper is three times the probability of not hitting the target on a stormy day due to high wind speed. The sniper fired two shots on the target on a stormy day when wind speed was very high. Find the probability that: (i) target is hit (ii) atleast one shot misses the target.
Show Hint
When evaluating phrases like "at least one", always consider using the complement rule: \(P(\text{At least one event happens}) = 1 - P(\text{None of the events happen})\). This simplifies the calculation significantly by reducing multiple cases down to a single scenario.
Concept:
Let the event of hitting the target in a single shot be denoted as \(H\), and the event of missing the target (not hitting) be denoted as \(M\) or \(H'\). Since these two outcomes are mutually exclusive and collectively exhaustive for any single attempt, the sum of their probabilities must equal \(1\):
\[
P(H) + P(M) = 1
\]
For independent repetitions of an experiment (like firing multiple distinct shots), the joint probability of independent events is the product of their individual probabilities.
• The probability that the target is hit at least once in \(n\) shots can be computed using the complement rule:
\[ P(\text{target is hit}) = 1 - P(\text{all shots miss}) \]
• The event that "at least one shot misses" is the complement of the event that "all shots hit the target".
Step 1: Determine the individual probabilities of hitting and missing.
Let the probability of not hitting the target (missing) be \(P(M) = p\).
According to the problem description, the probability of hitting the target, \(P(H)\), is three times the probability of missing. Therefore, we can write:
\[
P(H) = 3 \cdot P(M) = 3p
\]
Since the total probability of all possible outcomes in a single trial is always equal to \(1\):
\[
P(H) + P(M) = 1
\]
Substituting the expressions in terms of \(p\):
\[
3p + p = 1
\]
\[
4p = 1 \quad \Rightarrow \quad p = \frac{1}{4}
\]
Thus, we have the individual probabilities for each single shot:
\[
P(M) = \frac{1}{4}
\]
\[
P(H) = 3\left(\frac{1}{4}\right) = \frac{3}{4}
\]
Step 2: Find the probability that the target is hit (i.e., at least one shot hits).
The sniper fires two independent shots. Let us denote the outcomes of the first and second shots as pairs. The target is considered "hit" if either the first shot hits, the second shot hits, or both shots hit.
Using the complement method, the target is hit if it does not happen that both shots miss:
\[
P(\text{target is hit}) = 1 - P(\text{both shots miss})
\]
Since the two shots are independent events, the probability that both miss is:
\[
P(\text{both shots miss}) = P(M \text{ on shot 1}) \times P(M \text{ on shot 2})
\]
\[
P(\text{both shots miss}) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}
\]
Now, substituting this back into our complement formula:
\[
P(\text{target is hit}) = 1 - \frac{1}{16} = \frac{16 - 1}{16} = \frac{15}{16}
\]
Step 3: Find the probability that at least one shot misses the target.
The phrase "at least one shot misses" means that either the first shot misses, the second shot misses, or both shots miss. The only outcome not included in this description is when both shots successfully hit the target.
Using the complement framework once again:
\[
P(\text{atleast one shot misses}) = 1 - P(\text{both shots hit})
\]
Since the shots are independent, the probability that both shots hit the target is:
\[
P(\text{both shots hit}) = P(H \text{ on shot 1}) \times P(H \text{ on shot 2})
\]
\[
P(\text{both shots hit}) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}
\]
Now, substituting this value back into the complement equation:
\[
P(\text{atleast one shot misses}) = 1 - \frac{9}{16} = \frac{16 - 9}{16} = \frac{7}{16}
\]
