Question

A hunter is firing at a target. He has only 10% chance of hitting it in one round. The number of rounds he must fire in order to have at least 50% chance of hitting the target at least once, is

• A. 8
• B. 7
• C. 6
• D. 5

Hint:

When dealing with "at least one" probability problems, it is almost always easier to calculate the probability of the complement event (hitting zero times) and subtract it from 1.

Answer 1

The Correct Option is B

Step 1: Identify single-shot probabilities.
Each round hits with probability $p=0.1$, so it misses with probability $q=1-p=0.9$.
Step 2: Probability of missing every round.
Shots are independent, so missing all $n$ rounds has probability $q^n=(0.9)^n$.
Step 3: Probability of at least one hit.
The complement of all-misses is at least one hit, with probability $1-(0.9)^n$.
Step 4: Set up the requirement.
We need this to be at least $50\%$: \[ 1-(0.9)^n\ge0.5\ \Longrightarrow\ (0.9)^n\le0.5. \]
Step 5: Test increasing $n$.
$(0.9)^5=0.59049>0.5$; $(0.9)^6\approx0.5314>0.5$; $(0.9)^7\approx0.4783<0.5$.
Step 6: Pick the smallest valid $n$.
The condition first holds at $n=7$, so he must fire $7$ rounds. This is option 2.
\[ \boxed{n=7} \]