# binomial probability distribution question #3 It is known that 20% of all persons given a certain medication get very drowsy within 2 minutes. Find the probabilities that among 15 persons given...

**binomial probability distribution question #3**

It is known that 20% of all persons given a certain medication get very drowsy within 2 minutes. Find the probabilities that among 15 persons given the medication, at least 5 will get very drowsy within 2 minutes.

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You need to evaluate binomial probability to find what is the probability that among 15 persons given a certain medication, at least 5 will get very drowsy within 2 minutes, hence you need to use the formula:

`P(r) = C_n^r*p^r*q^(n-r)`

n denotes the number of persons.

p denotes the probability for persons that take a certain medication to get very drowsy within 2 minutes.

q denotes the probability for persons that take a certain medication not to get drowsy within 2 minutes.

p+q = 1

r = 5

The problem provides the value of `p = 0.2 (20%), ` hence you may find the value of q such that: `q = 1 - p =gt q = 1 - 0.2 = 0.8`

`` You may evaluate the binomial probability that among 15 persons given the medication, at least 5 will get very drowsy within 2 minutes such that:

`P(5) = C_15^5*(0.2)^5*(0.8)^(15-5)`

Using the factorial formula for binomial coefficient yields:

`C_15^5 = (15!)/(5!*10!)=gtC_15^5 = (10!*11*12 *13*14*15)/(1*2*3*4*5*10!)=gtC_15^5 = 11*13*21*15 = 45045`

`P(5) = 45045*0.00032*0.32768*0.32768*0.8*0.8 `

`P(5) = 45045*0.00032*0.10737 = 0.1389`

**Hence, evaluating the probability that among 15 persons given a certain medication, at least 5 will get very drowsy within 2 minutes yields P(5)= 13.9%.**