The probability of a shooter hitting a target is $\frac{2}{3}$. Minimum how many number of times must be shooter fire so that the probability of hitting the target at least once is more than 0.98.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
P(hit target atleast once)
let n → no of times
1 - P(no hit) = $1-{^nC}_0(\frac{1}{3})^n≥0.98$
for least n
so $0.02=\frac{1}{3^n}$
so $3^n=50$
so $n≥4$