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In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data were compiled into the following table.

$$\begin{array}{|c|c|c|c|}\hline \text { Shirtt } & { \leq 210} & {211-250} & {251-290} & {>290} \\ \hline 1-3 & {21} & {5} & {0} & {0} \\ \hline 34-66 & {6} & {18} & {7} & {4} \\ \hline 66-99 & {6} & {12} & {22} & {5} \\ \hline\end{array}$$

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

a. Find the probability that his shirt number is from 1 to 33.

b. Find the probability that he weighs at most 210 pounds.

c. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210

a) $\frac{26}{106}$

b) $\frac{33}{106}$

c) $\frac{21}{106}$

d) $\frac{38}{106}$

e) $\frac{21}{33}$

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Hi, we're answering Question 1 13 that looks at the weights and the jersey numbers of the members of the San Francisco 40 Niners and the Dallas Cowboys. So you have table 3.23 in your book that you can look at and starting with her, eh? Probably that a shirt is numbered from 1 to 33. So we're going to look at the row that has 1 to 33. So there's 21 plus five plus zero plus zero on the top. So those are favorable outcomes because all of those people have those and then for the bottom were actually going to add up all of the values in our table. Eso What we're doing is adding the 21 I'll put this on the top, because will you be using this again? So we have 21 plus the five. Um, we could add in the zeros overly will make any difference in our totals. Um, so that's the top row. And then the six and the 18 and the seven in the four for the second row. And then finally, for our third round a six, the 12 the 22 and the five that gives us a total of 106. So our total number of possible outcomes is 106 And with this question, they leave the totals in our the probability sorry in, um, direction form. So we can actually just leave this as 26 over 106 Part B s is to find the win. The probably weighs at most £210. So at most means he needs to weigh less Stan or 210. So we would again be looking at the numbers that fit that which would just be the column of less than or equal to 210. So we would have 21 plus six plus six. And that's over all of our players, which is +106 So we get When we add those together, we get 33 out of 106 for part C and once the probably that the shirt number is between one and 33 they wait most £210. So we want to find the road that has the shorts. 13 33. The column that has less than or equal to £210. So that gives us 21 out of our total number, which is 106 And we're gonna want to keep that 106 in the back of her mind as I flipped to the next slide, which is asking us for the shirts 13 33 or at most, £210. So this one is still gonna be out of 106 So since I remembering that from the last spring, I'm just gonna jot that down and then looking at our table, we need the shirts that air one through 33. So that would be the 21 plus the six plus the six plus the at most £210. What? We don't want to count the 21 that was the one through 33 twice, So we don't want to include those. So we're going to add just the five in the zero in the zero because there's not twice as many players that air one numbers one through 33 in less than 210. There's just the 21 I've already counted those right here, so I'm not going to add those in a second time. So that's going to give me a total of 38 out of 106 We could have also done this by taking the total off the one through 33 row, which would be 27 and 26. Sorry, I other grown and then adding the total from the column, which would be of less than or equal to 2 10 which is 33 and then subtracting the 21 that would we would be counting twice. We would still get the 38 out of 106 for that. And finally, the probably that the shirt is one through through 33. Give in. That little person weighs at most £210 so we have 21 shirts that are both one through 30 3 are 21 players that are both one through 33 way less than £210 and so that's given out of the most to £10. Were would add for the bottom the 21 plus the six plus the six that's in that column giving us 21 out of 33 for that probability,

Brigham Young University