Respuesta :
Solving the first problem: n=15 and p=0.4.
Explanation:
Step 1. We are given the number of trials n
[tex]n=15[/tex]And the probability of success:
[tex]p=0.4[/tex]We need to find the probability of 4 successes P(4).
Step 2. Since it s a problem where the outcome is either success or failure, we use the binomial distribution formula:
[tex]p(x)=\frac{n!}{(n-x)!x!}p^xq^{n-x}[/tex]Where p(x) is the probability of x successes. x is the number of successes, and q is the probability of a failure.
Step 3. In this case, x is 4:
[tex]x=4[/tex]The probability of failure is defined as follows:
[tex]\begin{gathered} q=1-p \\ \downarrow \\ q=1-0.4 \\ \downarrow \\ q=0.6 \end{gathered}[/tex]Step 4. Substituting the known values into the binomial distribution formula:
[tex]\begin{gathered} p(x)=\frac{n!}{(n-x)!x!}p^xq^{n-x} \\ \downarrow \\ p(4)=\frac{15!}{(15-4)!(4!)}(0.4)^4(0.6)^{15-4} \end{gathered}[/tex]Solving the operations:
[tex]\begin{gathered} p(4)=\frac{15!}{(11)!(4!)}(0.0256)(0.6)^{11} \\ \downarrow \\ p(4)=\frac{15!}{(11)!4!}(0.0256)(0.003628) \end{gathered}[/tex]Step 5. Solving step by step and simplifying the factorial expressions, the result is:
[tex]\begin{gathered} p(4)=\frac{15!}{(11)!4!}(0.0256)(0.003628) \\ \downarrow \\ p(4)=\frac{15\times14\times13\times12\times11!}{(11)!4!}(0.0256)(0.003628) \\ \downarrow \\ p(4)=\frac{15\times14\times13\times12}{4!}(0.0256)(0.003628) \\ \downarrow \\ p(4)=\frac{15\times14\times13\times12}{4\times3\times2\times1}(0.0256)(0.003628) \\ \downarrow \\ p(4)=\frac{32760}{24}(0.0256)(0.003628) \\ \downarrow \\ p(4)=(1365)(0.0256)(0.003628) \\ \downarrow \\ p(4)=0.126776 \end{gathered}[/tex]The probability is 0.126776
Answer:
[tex]P\left(4successes\right)=0.126776[/tex]