Respuesta :
Answer:
a) the probability mass function p(x) = P(X=x) is:
[tex]p(1)=0.5[/tex]
[tex]p(2)=0.333[/tex]
[tex]p(3)=0.166[/tex]
[tex]p(4)=0[/tex]
b) The expected number of calls E(X) you will have made when you talk to the first interested customer is 1.667 calls
Step-by-step explanation:
The probability that you will talk with one of the interest customers at the first call is:
[tex]p(1)=\frac{2}{4} =0.5[/tex]
Because, you have 4 possibles customers and 2 of them are interested.
Also, the probability that you will talk with one of the interest customers at the second call is:
[tex]p(2)=\frac{2}{4}*\frac{2}{3}=0.333[/tex]
Because, in the first call you are going to talk with one customer that it is not interested, so you have 4 possibles customers and 2 of then are not interested. Then, for the second call, you are going to talk with one customer that it is interested, so, now, you have 3 possibles customers and 2 of them are interested.
At the same way, the probability that you will talk with one of the interest customers at the third call is:
[tex]p(3)=\frac{2}{4}*\frac{1}{3}*\frac{2}{2}=0.166[/tex]
Finally, taking into account that there are 4 customers and 2 of them are interested, the maximum number of calls for find the first customer interested is 3, so P(4) is zero.
So, the probability mass function p(x) = P(X=x) is:
[tex]p(1)=\frac{2}{4}=0.5[/tex]
[tex]p(2)=\frac{2}{4}*\frac{2}{3}=0.333[/tex]
[tex]p(3)=\frac{2}{4}*\frac{1}{3}*\frac{2}{2}=0.166[/tex]
[tex]p(4)=0[/tex]
Then, the expected value is calculated as:
[tex]E(x)=x_1p(x_1)+x_2p(x_2)+... + x_np(x_n)[/tex]
Where [tex]x_1, x_2,...,x_n[/tex] are the possible values of the variable and [tex]p(x_1), p(x_2),...,p(x_n)[/tex] are their respectives probabilities.
Therefore, the expected number of calls that you will have made when you talk to the first interested customer is:
E(x)=1(0.5) + 2(0.333) + 3(0.166) + 4(0) = 1.667