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Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the probability that any particular couple or individual arrives late is .4 (a couple will travel together in the same vehicle, so either both people will be on time or else both will arrive late). Assume that different couples and individuals are on time or late independently of one another. Let X = the number of people who arrive late for the seminar.

a) Determine the probability mass function of X. [Hint: label the three couples #1, #2, and #3, and the two individuals #4 and #5.]

b) Obtain the cumulative distribution function of X, and use it to calculate P(2 <= X <= 6).

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Let $L_i$ be an indicator function, equal to 1 if #i is late. Then $$ X = 2 L_1 + 2 L_2 + 2 L_3 + L_4 + L_5 $$ where each $L_i$ is independent and follows Bernoulli distribution with probability $p=0.4$. The easiest way to find probability mass function is through the probability generating function: $$ \mathcal{P}_X(s) = \mathbb{E}( s^X ) = \left(\mathcal{P}_L(s^2)\right)^3 \cdot \left(\mathcal{P}_L(s)\right)^2 = (1-p+p s^2)^3 (1-p + p s)^2 $$ Now expand this polynomial. Then $$ \mathbb{P}(X= k) = [s]^k \mathcal{P}_X(s) $$

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We can patiently work out the probability $f_(X)(0)$ that exactly $0$ people arrive late, the probability $f_X(1)$ that exactly $1$ person arrives late, the probability $f_X(2)$ that exactly $2$ people arrive late, and so on up to the probability $f_X(8)$ that $8$ people arrive late.

What is $f_X(0)$? The two singles must arrive on time, and the three couples also must. It follows that $f_X(0)=(0.6)^5$.

What is $f_X(1)$? Exactly $1$ person, a single, must arrive late, and all the rest must arrive on time. The tardy single can be chosen in $2$ ways. The probabiliy that (s)he arrives late is $0.4$. The probability that the other single and the three couples arrive on time is $(0.6)^4$. It follows that $$f_X(1)=(2)(0.4)(0.6)^4.$$

What is $f_X(2)$? Two late can happen in two different ways. Either (i) the two singles are late, and the couples are on time or (ii) the singles are on time but one couple is late.

(i) The probability that the two singles are late, but the couples are not is $(0.4)^2(0.6)^3$.

(ii) The probability that the two singles are on time is $(0.6)^2$. Given that the singles are on time, the late couple can be chosen in $3$ ways. The probability that it is late is $0.4$, and the probability the other two couples are on time is $(0.6)^2$. So the probability of (i) is $(0.6)^2(3)(0.4)(0.6)^2$, which looks better as $(3)(0.4)(0.6)^4$. It follows that $$f_X(2)=(0.4)^2(0.6)^3+(3)(0.4)(0.6)^4.$$

What is $f_X(3)$. Here a single must arrive late, and also a couple. The late single can be chosen in $2$ ways, The probability (s)he is late but the other single is not is $(0.4)(0.6)$. The late couple can be chosen in $3$ ways. The probability it is late and the other two couples are not is $(0.4)(0.6)^2$. Putting things together, we find that $$f_X(3)= (2)(3)(0.4)^2(0.6)^3.$$

What is $f_X(4)$? This calculation is like the one for $f_X(2)$, since we either have the two singles and one couple late, or two couples late. So the calculation will break up into two cases.

Since this is homework, it's now your turn. Some, like $f_X(8)$, will be very easy. Some will be less easy, but they all follow the lines of the calculations above. It is quite difficult (at least for me) to do this sort of thing without making at least one slip. So when you find all the answers, add them up and see whether the sum is $1$, as it must be.

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If you can use excel simply use the binomial distribution function: BINOM.DIST(x,n,p,cumulative).

Where x is the number of late people, n is the total number of people, p is the probability of someone being late, and cumulative is a bool that says whether or not to calculate the cumulative probability. Consider couples to be one person, since they arrive together.

So for the pmf use BINOM.DIST with cumulative set to false on each of the x values, then to find the cdf repeat with cumulative set to true.

This will work up to 5 then you have to use some logic to find the rest from the values you already have.

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"Consider couples to be one person" – No. Also, a software answer does not seem appropriate for this question. – epimorphic Oct 14 '14 at 3:10
A single and a couple can both be considered one random variable since the couples share the outcome of either being on time or late. Do it yourself the solution is correct up to 5 where you then need extra logic to consider the different permutations of singles and couples that could result in 6-8. But up to 5 it's just a binomial distribution problem. And an excel solution is perfectly appropriate for a statistics problem. My entire statistics class was in Excel, even the tests. – Alec Karfonta Oct 14 '14 at 3:20
No, it's not a simple binomial distribution problem for $x \leq 5$. Please see the other answers to this question. Check: $\mathtt{BINOM.DIST}(1,5,0.4,\mathtt{false}) = (\boldsymbol 5)(0.4)(0.6)^4$, while André Nicolas explains how the probability of exactly one person being late is actually $(\boldsymbol 2)(0.4)(0.6)^4$. The discrepancy is due to the fact that of the five possibilities where one group is late, only two actually correspond to one person being late. In the other three, two people (one couple) are late. – epimorphic Oct 14 '14 at 5:46

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