Question

Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X=17), n=18, p=0.9

Answer 1

Answer:

P(X = 17) = 0.3002

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 18, p = 0.9$

We want P(X = 17). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 17) = C_{18,17}.(0.9)^{17}.(0.1)^{1} = 0.3002$

P(X = 17) = 0.3002

Answer 2

Answer:

P(X=17) = 0.3002 .

Step-by-step explanation:

We are given that the random variable X has a binomial distribution with the given probability of obtaining a success.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 18

            r = number of success = 17

           p = probability of success which in our question is given as 0.9 .

So, X ~ $Binom(n=18,p=0.9)$

We have to find the probability of P(X = 17);

P(X = 17) = $\binom{18}{17}0.9^{17} (1-0.9)^{18-17}$

               = $18 \times 0.9^{17} \times 0.1^{1}$    { $\because \binom{n}{r} = \frac{n!}{r! \times (n-r)!}$ }

               = 0.3002

Therefore, P(X=17) = 0.3002 .