When you ask a student of the sample if he/she knows the channel there are only two possible outcomes: yes or no. This means that we are dealing with a binomial probability distribution problem. Asking a student if he/she knows the channel is considered a "trial", the number of students in the sample is the number of trials that we'll name n, an student answering that he/she knows the channel will be considered a "success" and the number of successes in the sample (i.e. the number of students from the sample knowing the channel) will be k. We know that the probability that a student knows the channel is 0.6 (this is 60% written as a fraction), this is the probability of success in a single trial and we are going to use p to refer to it.
For a binomial probability distribution the probability of having k successes in n trials with a single trial probability p is given by the following formula:
[tex]P(k,n)=\frac{n!}{(n-k)!k!}\cdot p^k\cdot(1-p)^{n-k}[/tex]The probability that at most 3 student out of the total 15 students in the sample know the channel is given by the sum of 4 probabilities:
- The probability that no student knows the channel P(0,15).
- The probability that 1 student knows the channel P(1,15).
- The probability that 2 students know the channel P(2,15).
- The probability that 3 students know the channel P(3,15).
So we have to find each of these probabilities and then we add them. We begin with P(0,15):
[tex]P(0,15)=\frac{15!}{(15-0)!0!}0.6^0(1-0.6)^{15-0}=1\cdot1\cdot0.4^{15}=0.000001[/tex]Then P(1,15):
[tex]P(1,15)=\frac{15!}{(15-1)!1!}0.6^1(1-0.6)^{15-1}=15\cdot0.6\cdot0.4^{14}=0.00002[/tex]And P(2,15):
[tex]P(2,15)=\frac{15!}{(15-2)!2!}0.6^2(1-0.6)^{15-2}=105\cdot0.6^2\cdot0.4^{13}=0.00025[/tex]And P(3,15):
[tex]P(3,15)=\frac{15!}{(15-3)!3!}0.6^3(1-0.6)^{15-3}=455\cdot0.6^3\cdot0.4^{12}=0.00165[/tex]Then we add the four probabilities:
[tex]\begin{gathered} P(0,15)+P(1,15)+P(2,15)+P(3,15)=0.000001+0.00002+0.00025+0.00165 \\ \begin{equation*} 0.000001+0.00002+0.00025+0.00165=0.001921 \end{equation*} \end{gathered}[/tex]So the probability is 0.001921.
In part e we must find the probability that less than 3 know the channel. This is given by the sum of P(0,15), P(1,15) and P(2,15) so we get:
[tex]P(0,15)+P(1,15)+P(2,15)=0.000001+0.00002+0.00025=0.000271[/tex]AnswerWe must round the answers to 4 decimal places so the answers are:
c. 0.0019
e. 0.0003
