Question

Given the number of trials and the probability of success, determine the probability indicated: (Hint use binomial distribution formula use factorials ! in showing your work)n = 15, p = 0.4, find P(4 successes) n = 12, p = 0.2, find P(2 success ) n = 20, p = 0.05, find P(at most 3 successes) (hint for c. P (at most 3 successes) = P(x ≤3)= P(x= 0) + P(x = 1)+ P(x = 2)+ P(x = 3)

Answer 1

Solving the first problem: n=15 and p=0.4.

Step-by-step explanation:

Step 1. We are given the number of trials n

`n=15`

And the probability of success:

`p=0.4`

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:

`p(x)=(n!)/((n-x)!x!)p^xq^(n-x)`

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:

`x=4`

The probability of failure is defined as follows:

`\begin{gathered} q=1-p \\ \downarrow \\ q=1-0.4 \\ \downarrow \\ q=0.6 \end{gathered}`

Step 4. Substituting the known values into the binomial distribution formula:

`\begin{gathered} p(x)=(n!)/((n-x)!x!)p^xq^(n-x) \\ \downarrow \\ p(4)=(15!)/((15-4)!(4!))(0.4)^4(0.6)^(15-4) \end{gathered}`

Solving the operations:

`\begin{gathered} p(4)=(15!)/((11)!(4!))(0.0256)(0.6)^(11) \\ \downarrow \\ p(4)=(15!)/((11)!4!)(0.0256)(0.003628) \end{gathered}`

Step 5. Solving step by step and simplifying the factorial expressions, the result is:

`\begin{gathered} p(4)=(15!)/((11)!4!)(0.0256)(0.003628) \\ \downarrow \\ p(4)=(15*14*13*12*11!)/((11)!4!)(0.0256)(0.003628) \\ \downarrow \\ p(4)=(15*14*13*12)/(4!)(0.0256)(0.003628) \\ \downarrow \\ p(4)=(15*14*13*12)/(4*3*2*1)(0.0256)(0.003628) \\ \downarrow \\ p(4)=(32760)/(24)(0.0256)(0.003628) \\ \downarrow \\ p(4)=(1365)(0.0256)(0.003628) \\ \downarrow \\ p(4)=0.126776 \end{gathered}`

The probability is 0.126776

Answer:

`P\left(4successes\right)=0.126776`