There are 3 friends of Michael: Tommy, Arthur, John. Each friend decides whether or not they want to come to lunch in his home restaurant following the socialdistancing norm, independently of the others. Tommy wants to come with a probability 0.8. Arthur wants to come with a probability 0.6. John wants to come with a probability 0.1. Unfortunately, if all 3 of them want to come, he must turn one of them away. Since the table capacity is 2 guests. Otherwise, he will take everyone that comes. Let X be the number of friends that Michael serves at lunch. What is the expectation E[X]? (a) 1.15 (b) 1.38 (c)1.45 (d)0.95
given solution :The sample space is$S_X = {0, 1, 2}$. since we can have anywhere from 0 to 2 people. By independence, $P(X = 0) = P(T ^C, A^C, J^C) = P(T ^C)P(A ^C)P(J ^C) = (0.2)(0.4)(0.9) = 0.072\\ P(X = 1) = P(T, A^C, J^C) + P(T ^C, A, J^C) + P(T ^C, A^C, J) = 0.404\\ P(X = 2) = 1 − P(X = 0) − P(X = 1) = 0.524\\ E[X] = \sum i.PX(i) = 1.452 $
my solution :$0(0.072) +1(0.404)+ 2(0.476) + 2(\frac{3}{8}) 0.048=1.392$
where $0.476$ is probability (exactly 2 people appeared)
$\frac{3}{8}$ = probability(2 people served /3 people appeared)
please tell where I am wrong
Your mistake is multiplying the last term by $\frac{3}{8}$. Rest is correct. Please note that $0.048$ is the probability of all three of them appearing for lunch and it is certain that if all three appear, only two will be served. That is why the book solution considers cases $ \small X = 0, X = 1$ and $ \small X \geq 2$.
$ \small P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 0.524$, which is same as $ \small P(X = 2) + P(X = 3) = 0.476 + 0.048$