• Given:
[tex]\begin{gathered} n=15 \\ p=0.4 \end{gathered}[/tex]You need to find this probability:
[tex]P(x=4)[/tex]The Binomial Distribution Formula is:
[tex]P(x)=\frac{n!}{(n-x)!x!}p^x(1-p)^{n-x}[/tex]Where "n" is the number of trials, "x" is the number of successes desired, and "p" is the probability of getting a success in one trial.
In this case you know that:
[tex]x=4[/tex]Therefore, by substituting values into the formula and evaluating, you get:
[tex]P(x=4)=(\frac{15!}{(15-4)!4!})(0.4)^4(1-0.4)^{15-4}[/tex][tex]P(x=4)=(\frac{15!}{(11)!4!})(0.4)^4(0.6)^{11}[/tex][tex]P(x=4)\approx0.1268[/tex]• Given that:
[tex]\begin{gathered} n=12 \\ p=0.2 \end{gathered}[/tex]You need to find:
[tex]P(x=2)[/tex]In this case:
[tex]x=2[/tex]Therefore, using the same formula, you get:
[tex]P(x=2)=(\frac{12!}{(12-2)!2!})(0.2)^2(1-0.2)^{12-2}[/tex][tex]P(x=2)=(\frac{12!}{(10)!2!})(0.2)^2(0.8)^{10}[/tex][tex]P(x=2)\approx0.2835[/tex]• Given:
[tex]\begin{gathered} n=20 \\ p=0.05 \end{gathered}[/tex]You need to find:
[tex]P(x\leq3)[/tex]This is:
[tex]P(x\leq3)=P(x=0)+P(x=1)+P(x=2)+P(x=3)[/tex]Therefore, using the formula, set up:
[tex]P(x\leq3)=(\frac{20!}{(20-0)!0!})(0.05)^0(1-0.05)^{20}(\frac{20!}{(20-1)!1!})(0.05)^1(1-0.05)^{20-1}+(\frac{20!}{(20-2)!2!})(0.05)^2(1-0.05)^{20-2}+(\frac{20!}{(20-3)!3!})(0.05)^3(1-0.05)^{20-3}[/tex]Then, you get:
[tex]P(x\leq3)\approx0.9841[/tex]Hence, the answers are:
[tex]P(x=4)\approx0.1268[/tex][tex]P(x=2)\approx0.2835[/tex][tex]P(x\leq3)\approx0.9841[/tex]
