Question

Need to find 2 successes and 4. Successes probablity first 2

Answer 1

• Given:

`\begin{gathered} n=15 \\ p=0.4 \end{gathered}`

You need to find this probability:

`P(x=4)`

The Binomial Distribution Formula is:

`P(x)=(n!)/((n-x)!x!)p^x(1-p)^(n-x)`

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:

`x=4`

Therefore, by substituting values into the formula and evaluating, you get:

`P(x=4)=((15!)/((15-4)!4!))(0.4)^4(1-0.4)^(15-4)`
`P(x=4)=((15!)/((11)!4!))(0.4)^4(0.6)^(11)`
`P(x=4)\approx0.1268`

• Given that:

`\begin{gathered} n=12 \\ p=0.2 \end{gathered}`

You need to find:

`P(x=2)`

In this case:

`x=2`

Therefore, using the same formula, you get:

`P(x=2)=((12!)/((12-2)!2!))(0.2)^2(1-0.2)^(12-2)`
`P(x=2)=((12!)/((10)!2!))(0.2)^2(0.8)^(10)`
`P(x=2)\approx0.2835`

• Given:

`\begin{gathered} n=20 \\ p=0.05 \end{gathered}`

You need to find:

`P(x\leq3)`

This is:

`P(x\leq3)=P(x=0)+P(x=1)+P(x=2)+P(x=3)`

Therefore, using the formula, set up:

`P(x\leq3)=((20!)/((20-0)!0!))(0.05)^0(1-0.05)^(20)((20!)/((20-1)!1!))(0.05)^1(1-0.05)^(20-1)+((20!)/((20-2)!2!))(0.05)^2(1-0.05)^(20-2)+((20!)/((20-3)!3!))(0.05)^3(1-0.05)^(20-3)`

Then, you get:

`P(x\leq3)\approx0.9841`

Hence, the answers are:

`P(x=4)\approx0.1268`
`P(x=2)\approx0.2835`
`P(x\leq3)\approx0.9841`