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.

4

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 $n≥4$