contestada

Let X be a binomial random variable with n = 9 and p = 0.2 (nine trials, probability of success = 0.2). Find the probability of 4 successes, i.e., P(X = 4). Give your answer as a decimal to three decimal places, e.g., 0.123. Round as needed.

Respuesta :

Answer:

The probability of 4 successes,

P(X=4) =0.0660

Step-by-step explanation:

Step 1:-

Let X be a binomial random variable with n = 9 and p = 0.2

probability of successes p = 0.2

probability of failure q = 1-p = 1-0.2 = 0.8

By using binomial distribution

[tex]P(X=r) = n_{Cr} p^{r} q^{n-r}[/tex]

Step 2:-

The probability of 4 successes, i.e.,

[tex]P(X=4) = 9_{C4} (0.2)^{4} (0.8)^{9-4}[/tex]

by using [tex]n_{Cr} = \frac{n!}{(n-r)!r!}[/tex]

[tex]9_{C4} = \frac{9!}{(9-4)!4!} = \frac{9X8X7X6X5!}{5!4!}[/tex] [tex]= \frac{9X8X7X6}{4!} = \frac{9X8X7X6}{4X3X2X1} =126[/tex]

[tex]P(X=4) =126 X (0.0016)(0.32768)[/tex]

P(X=4) =0.0660

conclusion:-

The probability of 4 successes, that is P(X=4) =0.0660

Otras preguntas