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 ≤ 3), n = 5, p = 0.2

Answer 1

Answer:

P(X ≤ 3) = 0.9933.

Step-by-step explanation:

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

Also, given n = 5, p =0.2.

The above situation can be represented through binomial distribution;

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

where, n = number of samples (trials) taken = 5

r = number of success = less than equal to 3

p = probability of success which in our question is 0.20.

Let X =  A random variable

So, X ~ Binom(n = 5, p = 0.20)

Now, the probability that X is less than and equal to 3 is given by = P(X ≤ 3)

P(X ≤ 3)  = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

=  $\binom{5}{0} \times 0.20^{0} \times (1-0.20)^{5-0}+\binom{5}{1} \times 0.20^{1} \times (1-0.20)^{5-1}+\binom{5}{2} \times 0.20^{2} \times (1-0.20)^{5-2}+\binom{5}{3} \times 0.20^{3} \times (1-0.20)^{5-3}$

 =  $1 \times 1 \times 0.80^{5}+5 \times 0.20^{1} \times 0.80^{4}+10 \times 0.20^{2} \times 0.80^{3}+10 \times 0.20^{3} \times 0.80^{2}$  

=  0.9933