Combinatorics Issues

I have three homework questions, all of which have the mutual problem of combinatorics

The first one:

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday’s mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one $P(Wed.)=3$, $P(Thurs.)=.4$, $P(Fri.)=.2$, and $P(Sat.)=.1$. Let Y= the number of days beyond Wednesday that it takes for both magazines to arrive (so possible Y values are 0, 1, 2, or 3). Compute the pmf of $Y$. [Hint: There are 16 possible outcomes; $Y(W,W) = 0$, $Y(F, TH) = 2$, and so on.]

Now, I know that the explicitly tell you the number of outcomes in the sample space, and it is small enough that you could generate the sample space $S$; but I was wondering, are there any fancy combinatoric techniques that you could use to count them? I have rather poor skills when it comes to combinatorics.

The second question is:

A new battery’s voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that 90% of all batteries have acceptable voltages. Let Y denote the number of batteries that must be tested.

a.What is p(2), that is,$P(Y=2)$?

b.What is p(3)? [Hint:There are two different outcomes that result in $Y=3$]

c. To have $Y=5$ , what must be true of the fifth battery selected? List the four outcomes for which $Y=5$ and then determine $p(5)$.

d.Use the pattern in your answers for parts (a)–(c) to obtain a general formula for p(y).

For this particular question, I am working on part c). I know for part c), that the 5th battery test in the sequence must be an acceptable battery; if it occurred any sooner in the sequence, then would wouldn't have to test 5 batteries. So, the last battery tested is fixed; that is _ _ _ _ A, where the first four slots can take on one other A (acceptable and 3 different U (unacceptable).

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