Question

Consider 3 trials, each having the same probability of successes. Let X denote the total number of successes in these trials. If E(X) = 1.8, what is:

a. the largest possible value of P{X=3}?
b. the smallest possible value of P{X=3}?

In both cases, construct a probability scenario that results in P{X=3} having the stated value.

Would really appreciate some help on this one guys...I'm told the answer to A is 0.6 but why?

Answer 1

they each have the same probability right?
\[E(x)=P(x=0)\times 0+P(x=1)\times 1+P(x=2)\times 2+P(x=3)\times 3=1.8\]

Answer 2

if all the first ones are zero, then
\[P(x=3)\times 3=1.8\] so \[P(x=3)=0.6\]

Answer 3

See that's interesting because that emulates perfectly a lot of the solutions. My only question (stupidly) is why:

P(X=0)*0 + P(X=1)*1 ....
Is that because the value of that probability i.e. take P(X=1) for example, is 1?

Answer 4

Wait yes my question is why are the first ones zero.

How do you simply jump to P(x=3)*3 = 1.8 ?