# binomial probability distribution question #1If the probability is 0.60 that a person traveling in a certain airline will pay extra to see a movie, what is the probability that only 3 of 6 persons...

binomial probability distribution question #1

If the probability is 0.60 that a person traveling in a certain airline will pay extra to see a movie, what is the probability that only 3 of 6 persons traveling on this airline will pay an extra to see a movie?

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You need to evaluate the binomial probability, hence you need to use the following formula:

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

You need to know what the letter n,p,q,r denote, hence:n denotes the number of trials, p denotes the probability that the event will happen, q expresses the probability that the event will not happen, r denotes the number of wanted events.

You need to remember that `p+q=1` .

The problem provides the information about the probability that a person traveling in a certain airline will pay extra to see a movie, hence `p=0.6` .

Since the value of p is given, you may evaluate the probability that a person traveling in a certain airline will not pay extra to see a movie such that:

`q = 1 - p =gt q = 1 - 0.6 =gt q = 0.4`

You may evaluate the the probability that only 3 of 6 persons traveling on this airline will pay an extra to see a movie such that:

`P(3) = C_6^3*(0.6)^3*(0.4)^(6-3)`

Using the factorial formula for combination yields: `C_6^3 = (6!)/(3!*3!) =gt C_6^3 = 4*5*6/6 = 20`

`P(3) = 20*0.216*0.064=0.27648`

**Hence, the probability that only 3 of 6 persons traveling on this airline will pay an extra to see a movie is of 27.648%.**