Then I guess the problem is solved.

Thank you.]]>

I am getting 1 for the total in the first table.

It seems to me you forgot to add the case when X=0.

]]>P(X)= Combination * (0.4^Total_Groups * 0.6^(5-Total_Groups))

And if I collapse rows with same number of people as a union of the two probabilities I receive:

Still no 1.0 in the total. What is wrong now?

]]>But once I recalculate whole table like this - the sum become 1.98.]]>

Here is a problem from the last homework:

Suppose we have three couples and two individuals - five independent groups in total (eight people). Three couples are marked #1, #2, and #3, groups of one person are marked #4 and #5. Each of these five groups can be late for a meeting with a probability 40%. All groups are independent from each other.

Let X be a number of people who arrived late.

Determine pmf and cdf for X.

Well, first I tried to formalize X:

Since there are two possible combinations for 2, 4, and 6 people, we need to take a union of probabilities for these combinations and final P(X) become this:

And from here I need to find

All examples in the textbook are just doing direct substitution p(x)=P(X=x) but if I summarize my P(X) I receive 1.4575616. And sum of p(x) for all x is supposed to be equal 1.0...

So, did I make a mistake somewhere or what am I missing?

]]>