Answer:
[tex]\approx 0.47[/tex]
Step-by-step explanation:
Let's define X = amount of customers that watch movies online. X has a binomial distribution with [tex]\frac{1}{4}[/tex] the probability of watching a movie online in a sample of 10 customers (we notate it as X [tex]\sim[/tex]B (10,[tex]\frac{1}{4}[/tex]), 10 and [tex]\frac{1}{4}[/tex] are the parameters of X).
Let's remember that the probability of getting exactly k successes (i.e. k customers that watch movies online) in n trials (n is the total of the sample), in a binomial distribution X with parameters (n, p), is given by the probability mass function:
[tex]P (X = k) = \binom{n}{k} * p^k * (1-p) ^(n-k)[/tex]
In our exercise, we need 3 or more but not greater than 7 successes. That means, we need to find out [tex]P(3\leq X\leq 7)[/tex].
Let's notice that this is equal to have 3, 4, 5, 6 or 7 customers that watch movies online.
So, [tex]P(3\leq X\leq 7) = P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)[/tex]
Now we just use the probability mass function, with n=10 and p=1/4, for each k ⊆ {3, 4, 5, 6, 7}:
[tex]\binom{10}{3} * 1/4^3 * (1-1/4)^(10-3) + \binom{10}{4} * 1/4^4 * (1-1/4)^(10-4) + \binom{10}{5} * 1/4^5 * (1-1/4)^(10-5) + \binom{10}{6} * 1/4^6 * (1-1/4)^(10-6) + \binom{10}{7} * 1/4^7 * (1-1/4)^(10-7) =[/tex]
[tex]\frac {10!} {3!7!} * 1/4^3 * 3/4^7 + \frac {10!} {4!6!} * 1/4^4 * 3/4^6 + \frac {10!} {5!5!} * 1/4^5 * 3/4^5 + \frac {10!} {6!4!} * 1/4^6 * 3/4^4 + \frac {10!} {7!3!} * 1/4^7 * 3/4^3 =[/tex]
Now we just do the calculation and we get
[tex]P(3\leq X\leq 7) \approx 0.47[/tex]
