Given:
[tex]\begin{gathered} Probability(success)=0.303 \\ Number-of-trials=6 \end{gathered}[/tex]To Determine: The probability of at least 3 hits
Solution
This is a binomial distribution case
[tex]P(X\ge3)=1-P(X\leq2)[/tex]The formula for finding binomial expansion is
From the given
[tex]\begin{gathered} n=6 \\ x=3 \\ p=0.303 \\ q=1-0.303=0.697 \end{gathered}[/tex]Using calculator
[tex]\begin{gathered} P(x\ge3)=0.261266 \\ \approx0.261 \end{gathered}[/tex]