# Spring 2013 – MAT 1272 Statistics – Reitz

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• Questions about WeBWorK, classwork, and exam reviews
• #13096

If you have a question about something on WeBWorK, something we did in class, or a review for the exam, go ahead and post it here. If someone asks a question and you know the answer, please respond to them!

#16505

Hint for WeBWorK #4, problem #3: this is by far the trickiest problem on this assignment (in my opinion — let me know if you think differently). If you’re stuck, take a look in the book at Section 3.1 Examples 4 (page 131) and 11 (page 137). If you’re still stuck, post something here!

#16597

how do i solve the problem for number 1, when it says:

Two coins are tossed.

and some of the questions that an “And”, or two of the same kind or “one” of one kind
are a bit tricky.

#16598

and i’m still trying number 3,

with just a slight problem of the second part of number 6 and 7 since both are related.

#16599

@Anil, For problem 1, first write out the sample space (that is, make a list of all possible outcomes when you flip two coins) — there should be 4 altogether. For each of the parts of the problem, figure out how many outcomes satisfy that part — then make a fraction to get the probability. For example, for part 1 “One head and one tail”, how many outcomes have one head and one tail? Put that number on top, and the total number of possible outcomes on the bottom, to get the probability. Use a similar strategy for parts 2,3 and 4. If you’re still stuck, write back and give me a list of the outcomes you’re considering.

For problem #3, keep in mind that if I tell you the first letter MUST be ‘K’, then you don’t have any choices there — so for the first spot you’ll multiply by 1 (instead of by 10 or 9).

Hope this helps

#16600

ok..i think i understand,…….i’ll also read the text book.

oh about the coin, i know it should be 4, but does the second coin toss occur after the first coin toss…. as in the attached file*

#16601

@Anil, regarding the coins, your picture is right – you can think of the coin tosses happening one after another.

#16604

i just finally got through the reaming problems… the book helped with the second part of #6 and #7 for the “And” when i used

P (A and B) = P(A)*P(A | B)

and i finally got the correct set of possible outcomes of coins toss for #1 since i noticed the tree diagram was different than i would have drawn it.

and i finally got the Letters and digits understood. for #3….with the class work and text book.

Thanks.

#16634

Hi Professor Reitz,

I have spent 2 hours on 2 questions, it’s been unproductive and I have 5 other classes and can’t figure out the method to solution.

Any problems similar to this 1. Given that , , and , find the joint probability P(A and B):

And the remaining homework.

Of the 12 people who show up, 2 are women. How many ways are there to choose 10 players to take the field if at least one of these players must be women?

Of the 12 people who show up, 2 are women. How many ways are there to choose 10 players to take the field if at least one of these players must be women?

Can you show me how to do it?

#16635

Hi Andrew,

1. The first problem you mention didn’t copy/paste completely, but I took a look in the webwork — it looks like they give you the probabilities P(A), P(B), and P(B|A), and ask you to find the probability P(A and B). When you need to find the probability involving the word “and” — A and B — use the Multiplication Rule:
P(A and B) = P(A) * P(B|A)
so to get the answer, just multiply the two given probabilities P(A) and P(B|A). NOTE: You do NOT need the probability P(B) to answer this problem!

12. We need to choose 10 players out of 12 (NOTE: other students may have different numbers!). There are 2 women and 10 men, and we have to choose at least one woman. Think of it this way:
A. First, choose one woman (out of 2). That takes care of the first of our 10 choices, and it also guarantees that we now have at least one woman. Note that after we choose one woman, there are 11 people left to choose from.
B. So now we choose 9 more players, out of the remaining 11 people. We don’t worry whether they are women or men, since we already have one woman chosen.

Hope this helps – write back if you’re still stuck,
Mr. Reitz

#16639

I emailed you this but I’m not sure if it sent:

Hi professor,

I have attempted this problem now 26 times. I cannot seem to get the correct answer. It’s problem number 10.

Red is 10, blue marbles is 5 and white is 5 as well. The total is 20 and they’re asking if you draw 4 marbles at random

For the first part I did:
10C4*9c4*8c4*7c4 divided by 20c4
=13380.18576

The second part:
10c4*9c4*10c4*9c4 divided by 20c4
=144506.0062

The third part:
10c4*9c4*8c4*7c4 divided by 20c4
= 13380.18576

Thanks,

Sarina “holly” Roche

#16640

@sarina, I just replied to your email but I’ll copy the message here, as well, in case it is helpful to others (WebWork #5, problem 10):

You’re off to a good start. The total number of ways of choosing 4 marbles is, indeed, 20C4, and this goes on the bottom of the fraction in each part. Now we just have to figure out the top of the fraction:

Part 1: How many ways are there of choosing 4 marbles out of the 10 red marbles? This is all that’s needed on the top (your answer includes this, plus some additional stuff — it’s the additional stuff that’s throwing you off).

Part 2: We need to choose 2 red marbles, and then 2 non-red marbles.
– How many ways are there of choosing 2 red marbles (out of 10 total)?
– How many ways of choosing 2 non-red marbles (out of 10 total non-red marbles)?
Calculate these two separately and then multiply them together.

Part 3: Choose 4 non-red marbles (out of 10 total non-red marbles).

Hope this helps – write back if you’re still stuck.
Mr. Reitz

#16641

I thought this is how we did it in class

#16642

To choose 4 red marbles from 10 total, we simply use 10C4 (this takes care of all 4 marbles). Since we aren’t choosing any more marbles, we don’t need anything more than this on top.

(maybe you are mixing this up with other examples, in which we multiplied a series of fractions together like: 10/10 * 9/10 * 8/10 * 7/10 — we could maybe use a method like this if we were doing permutations, instead of combinations, but it’s much more efficient to get used to using the nCr and nPr formulas instead)

#16643

Part B I am now stuck on. I have 10c4 * 9c4 divided by 20c4 and it is incorrect

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