If his average is 0.196, the probability of hitting is also 0.196.
So, in a total of 9 bats, the probability of hitting at least 3 can be calculated by first calculating the probability of hitting 0, 1 or 2 times.
To do so, we can use the following formula:
[tex]P(x,9)=(9,x)\cdot p^x\cdot(1-p)^{9-x}[/tex]Where (9, x) is the binomial of 9 and x and p is the probability of hitting. So we have:
[tex]\begin{gathered} P(0,9)=(9,0)\cdot0.196^0\cdot0.804^9 \\ P=1\cdot1\cdot0.1404=0.1404 \\ \\ P(1,9)=(9,1)\cdot0.196^1\cdot0.804^8 \\ =9\cdot0.196\cdot0.1746=0.308 \\ \\ P(2,9)=(9,2)\cdot0.196^2\cdot0.804^7 \\ =36\cdot0.0384\cdot0.2172=0.3 \end{gathered}[/tex]So the probability of hitting at least 3 times can be calculated as:
[tex]\begin{gathered} P=1-(P(0,9)+P(1,9)+P(2,9)) \\ =1-(0.7484) \\ =0.2516 \end{gathered}[/tex]So the probability is 0.2516.