combinationspermutationsprobability
In how many ways can a team of 9 softball players be selected from 15 trying out, if 3 are experienced and….
b. Find the probability that there is at least one experienced player on the team.
c. Once the team is chosen, how many selections of 2 pitchers, 2 catchers, and 5
outfielders are possible?
I honestly have tried different things but I don't seem to be getting it.
You need to choose the captian so its only a way to choose the captain, for the defence you have 3 players so you have 3 choices and for the forward players you have 5 choices so the number of choices is : $1$x$5$x$3$=$15$ way or choice
Assuming that we do not care about labels (i.e. -1 DeAza(outfield), -2 Par (d.h.), -3 Young (outfield),... is considered to be the same outcome as -1 DeAza(d.h.), -2 Par (outfield), -3 Young (outfield),...), as Andre points out this is a perfect time to use the Addition Principle.
Our lineup will consist of 1 catcher, 3 outfielders, 4 infielders, and one additional person (the designated hitter) from one of those categories or from the pitcher category. Split it up into the four cases:
Notice, that there is absolutely no way that when we count scenarios from each of these cases, that we accidentally count the same scenario a second time in a different case. I.e., these events are mutually exclusive and therefore we can apply the addition principle and add the amount of possibilities from each individual case to get the number of possibilities total.
Let us examine what happens in one of the cases in closer detail.
This means that among our nine players in the batting lineup, two of them are catchers, 3 of them are outfielders, and 4 of them are infielders. We wish to count how many different lineups satisfy this statement. Here we will use the Multiplication Principle. Set up a sequence of questions/choices to determine which lineup we have.
Thus, by the multiplication principle there are $\binom{2}{2}\binom{6}{3}\binom{5}{4}\binom{12}{0}9!=36288000$ different outcomes for the first case (case A).
Calculate the number in case B, case C, and case D similarly (in your comment on the other copy you only used a $\binom{12}{1}$ for case D, however that only counts how many ways to pick one pitcher to be the designated hitter; you still need to pick the rest of the team and where they are in the lineup).
For part B, we can approach this two ways. The easier way I think is this: first ask the question "What is the probability that Jones is not picked to be in the starting lineup?" By answering this question, we are able to then easily answer the follow up question "What is the probability that Jones is in the starting lineup?".
We will answer the question of the probability that he is not exactly like how we answered for the first part of the question with the following adjustment: since we want to count the number of cases Jones is not in the lineup, redo the calculations for the first part as though Jones weren't on the team at all (I.e. remove Jones from the list of available outfielders, making it so that there are only $5$ outfielders available).
We again break it into the same cases as before: A) d.h. is a catcher, B) d.h. is an infielder, C) d.h. is an outfielder, D) d.h. is a pitcher.
Let us examine the first case in detail again:
For a total of $\binom{2}{2}\binom{5}{3}\binom{5}{4}\binom{12}{0}9!$ different outcomes for the first case. Calculate the other cases similarly.
Now that we have a tally of how many lineups do not have Jones on the roster, take the answer from the first half of the problem and subtract the number just calculated here to get the number of different ways that Jones IS on the starting lineup (as per a simple application of inclusion-exclusion ).
Take this newest number and divide by the answer found in the first half to get the probability that Jones is in the starting lineup as per the definition of probability in an equiprobable sample space: $$Pr(A) := \frac{|A|}{|S|}$$
Complete the problem by noting that, given that Jones is in the starting lineup, he is equally likely to be in each of the spots and apply the multiplication rule (probability).
$$Pr(\text{Jones bats}~4^{th}) = Pr(\text{Jones in lineup})\cdot Pr(\text{Jones bats}~4^{th}~|~\text{Jones in lineup})$$
There is another equally acceptable approach to solving the second part. Directly count the number of ways that Jones bats fourth. Do so by breaking it into cases:
Looking at one of these cases more closely, looking at A again: Break it up via multiplication principle as before. Since the D.H. is a catcher, that means our lineup will consist of 2 catchers, Jones, 2 other outfielders, and 4 infielders.
Do so similarly for the other cases to find the number of ways that Jones appears in the fourth batting position. Complete by applying the definition of probability (num outcomes for event / num outcomes regardless).
Best Answer
Part (b) is a case where the negative space is more easy to calculate: the probability that there are no experienced players.
The unconstrained choice has $\binom {15}{9}$ $ = \frac{\large 15!}{\large 9!6!} = \frac{\large 15\cdot14 \cdot13 \cdot12 \cdot11 \cdot10}{\large 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$ $ = 5005$ options.
The no-experienced-players choice has $\binom {12}{9} = \frac{\large 12!}{\large 9!3!} =\frac{\large 12 \cdot 11 \cdot 10}{\large 3 \cdot 2 \cdot 1} = 220$ options.
So the probability of having an experienced player on the team from a blind choice is $\frac{\large 5005-220}{\large 5005}$ $ = \frac{\large 4785}{\large 5005}$ $ = \frac{\large 87}{\large 91}$
Part (c) potentially depends on player skill mix, but absent any constraint will be $\binom {9}{2,2,5}$ $ = \binom 92\binom 72$