Question

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) at least one shot misses the target. 

Hint:

When two events are independent, the probability of both occurring is the product of their probabilities. Also remember that the probability of “at least one” event can often be calculated using the complement rule.

Answer 1

Step 1: Define probabilities.
Let the probability of hitting the target be \(p\).
Then the probability of missing the target is \[ 1 - p \] According to the question: \[ p = 3(1 - p) \] Step 2: Solve the probability equation.
\[ p = 3 - 3p \] \[ 4p = 3 \] \[ p = \frac{3}{4} \] Thus, \[ P(\text{hit}) = \frac{3}{4}, \quad P(\text{miss}) = \frac{1}{4} \] Step 3: Probability that the target is hit in two shots.
Since the sniper fires two shots and the shots are independent, the probability that both shots hit the target is \[ P(\text{both hits}) = \left(\frac{3}{4}\right)^2 \] \[ = \frac{9}{16} \] Step 4: Probability that at least one shot misses.
The probability that at least one shot misses is the complement of both shots hitting.
\[ P(\text{at least one miss}) = 1 - P(\text{both hits}) \] \[ = 1 - \frac{9}{16} \] \[ = \frac{7}{16} \] Step 5: Final conclusion.
Thus, the probability of both shots hitting the target is \(\frac{9}{16}\), and the probability that at least one shot misses the target is \(\frac{7}{16}\).
Final Answer:
\[ P(\text{target hit}) = \frac{9}{16}, \quad P(\text{at least one miss}) = \frac{7}{16} \]