Thursday

April 17, 2014

April 17, 2014

Number of results: 1,106

**probability/statistics**

a. (1/6+1/6) * (1/6+1/6+1/6) = ? b. same as a. c. Use same procedure. For either-or probabilities, add the individual probabilities. For both/all probabilities, multiply the individual probabilities.
*Monday, June 27, 2011 at 9:43am by PsyDAG*

**math**

For all probabilities, multiply individual probabilities. For either-or probabilities, add. (1/6 * 7/8) + (5/6 * 1/8) = ?
*Thursday, July 26, 2012 at 3:59pm by PsyDAG*

**statistics**

The either-or probability is found by adding the individual probabilities. The probabilities of each outcome need to add to one. What is the sum of your probabilities?
*Sunday, May 29, 2011 at 11:51pm by PsyDAG*

**problem stats**

To determine "either-or" probabilities, you add the probabilities of the individual events. Since the probabilities of each event are the same, you could multiply by 5. I hope this helps. Thanks for asking.
*Monday, October 15, 2007 at 8:29pm by PsyDAG*

**statistics**

For either-or probabilities, you add the individual probabilities. 4/52 + 13/52 - 1/52 = ? Subtract the 1/52, because one of the fours is a diamond. There are two black jacks in the deck. 2/52 = 1/26 To determine probabilities of all events occurring, multiply the individual ...
*Monday, June 20, 2011 at 5:45pm by PsyDAG*

**Statistics**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 2 correct = .2^2 * .8^4 = ? 1 correct = .2 * .8^5 = ? 3 correct = .2^3 * .8^3 = ? 4 correct = .2^4 * .8^2 = ? Either-or ...
*Sunday, March 3, 2013 at 7:22pm by PsyDAG*

**statistics**

This is the same as 1 or 2 people answering correctly. The probability of both/all events occurring is determined by multiplying the probabilities of the individual events. For 1, .4(.6)^7 = ? For 2. (.4)^2(.6)^6 = ? Either-or probabilities are found by adding the individual ...
*Saturday, December 8, 2012 at 1:39am by PsyDAG*

**Math**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. I assume you meant tails AND a five. P = 1/2 * 1/6 = ? Either-or probabilities are found by adding the individual probabilities.
*Friday, January 18, 2013 at 1:16pm by PsyDAG*

**Statistics**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 4/52 * 4/51 = ? Either-or probabilities are found by adding the individual probabilities. But it can be A,K or K,A, so multiply ...
*Tuesday, April 2, 2013 at 5:12pm by PsyDAG*

**Statistics**

It means that the first two times you get a 9 or less, but the third time you get a 10, 11 or 12. For third time, add the probabilities from Bobpursley. For either-or probabilities, add individual probabilities. (1/6*1/6) + (2*1/6*1/6) + (3*1/6*1/6) = ? Probability of 9 or ...
*Sunday, September 19, 2010 at 9:37pm by PsyDAG*

**Math - Probability**

3a) If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 3b) Either-or probabilities are found by adding the individual probabilities.
*Sunday, May 19, 2013 at 8:19pm by PsyDAG*

**Probability**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 1. .2 * (1-.3) = ? Either-or probabilities are found by adding the individual probabilities. 2. (.2 *3.) + (.8 *.3) = ?
*Thursday, February 20, 2014 at 7:06pm by PsyDAG*

**Probabilities**

Are you "answer mooching" or don't you understand anything about probabilities?
*Friday, May 1, 2009 at 10:38pm by Ms. Sue*

**math**

For either-or probabilities, add the individual probabilities. P(5) = 1/6 Sum of 5 can be obtained by (1,4)(4,1)(2,3)(3,2) P(1,4) = 1/6 * 1/6 = ? P(4,1) = ? P(3,2) = 1/6 * 1/6 = ? P(2,3) = ?
*Thursday, June 9, 2011 at 11:27pm by PsyDAG*

**math!**

3 = 1,2 or 2,1 = 2/36 7 = 6,1 or 1,6 or 5,2 or 2,5 or 3,4 or 4,3 = ? Either-or probabilities are found by adding the individual probabilities.
*Wednesday, July 31, 2013 at 9:20pm by PsyDAG*

**math**

Either-or probabilities are found by adding the individual probabilities.
*Saturday, July 3, 2010 at 11:24pm by PsyDAG*

**math**

Either-or probabilities are found by adding the individual probabilities.
*Sunday, April 28, 2013 at 7:34pm by PsyDAG*

**algebra**

Either-or probabilities are found by adding the individual probabilities. 1/2 + 1/2 = 1
*Friday, July 5, 2013 at 9:30pm by PsyDAG*

**algebra 2**

Less than 2 = 1 or 1/6 Greater than 3 = 4, 5 or 6 Either-or probabilities are found by adding the individual probabilities.
*Monday, April 29, 2013 at 2:02pm by PsyDAG*

**statistics**

Either-or probabilities are found by adding the individual probabilities. .23 + .05 = ?
*Wednesday, October 16, 2013 at 4:19pm by PsyDAG*

**statistics**

Either-or probabilities are found by adding the individual probabilities. .43 + .14 = ?
*Friday, February 28, 2014 at 8:56pm by PsyDAG*

**Math**

Can someone please help me. A single card is selected from a standard 52-card deck. B= the drawn card is black; R= the drawn card is red; Q= the drawn card is a queen; F= the drawn card is a face card ( a king, queen or jack). Without finding the probabilities, determine if B ...
*Saturday, December 1, 2012 at 3:17pm by James*

**Probability**

Can someone please help me. A single card is selected from a standard 52-card deck. B= the drawn card is black; R= the drawn card is red; Q= the drawn card is a queen; F= the drawn card is a face card ( a king, queen or jack). Without finding the probabilities, determine if B ...
*Saturday, December 1, 2012 at 7:34pm by James*

**Probability**

Can someone please help me. A single card is selected from a standard 52-card deck. B= the drawn card is black; R= the drawn card is red; Q= the drawn card is a queen; F= the drawn card is a face card ( a king, queen or jack). Without finding the probabilities, determine if B ...
*Sunday, December 2, 2012 at 2:00pm by Kelsey*

**math**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. P(1) = .85 * .15 = ? P(2) = .85 * .85 = ? Either-or probabilities are found by adding the individual probabilities.
*Wednesday, July 31, 2013 at 9:21pm by PsyDAG*

**Math**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 1. (1/5)^3 = ? 2. (4/5)^3 = ? 3. 1/5 4. 1/5 * 4/5 * 4/5 = ? 5. You do it, using similar process. 6. This is none good, one good ...
*Monday, May 6, 2013 at 5:38am by PsyDAG*

**statistics**

< 5cups = (1 - .24) Either-or probabilities are found by adding the individual probabilities.
*Monday, May 6, 2013 at 8:00pm by PsyDAG*

**7th Grade Math stuffs**

Compare probabilities of independent and probabilities of dependent events. Any examples???
*Thursday, January 17, 2013 at 10:41pm by lazyburrito*

**MATH**

Please do not use all caps. Not only is it harder to read, but it is like SHOUTING online. Thank you. Either-or probabilities are found by adding the individual probabilities. P (A or B) = 1/4 + 1/4 = ? P (1 or 2) = ? If the events are independent, the probability of both/all ...
*Wednesday, May 22, 2013 at 12:13am by PsyDAG*

**statistics**

You want the probability of 3, 4 or all 5 having part-time jobs. for 3 = .3*.3*.3*.7*.7 = ? for 4 = .3*.3*.3*.3*.7 = ? for 5 = ? For either-or probabilities, add the individual probabilities.
*Thursday, July 19, 2012 at 6:46am by PsyDAG*

**math**

So you are talking about 1,1; 1,2; 2,1; 2,2, 1,3; or 3,1. Each pair has a probability of 1/36. Either-or probabilities are found by adding the individual probabilities.
*Saturday, November 17, 2012 at 8:38pm by PsyDAG*

**Math - Probability (?)**

Are you choosing one or two? "boy or girl" vs. "both of them"? If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. Either-or probabilities are found by adding the individual ...
*Monday, May 13, 2013 at 1:08pm by PsyDAG*

**Statistics**

Probabilities only add to .90. Is the remaining .10 people who decided not to rent, or do you have a typo? 1. Either-or probabilities are found by adding the individual probabilities. 2. .16 + .12 + .24 = ? 3. If the events are independent, the probability of both/all events ...
*Monday, September 30, 2013 at 2:35pm by PsyDAG*

**Math**

Either-or probabilities are found by adding the individual probabilities. Add them up. P(5) on one die = 1/6 Sum = 5 = 4/36 1,4 4,1 2,3 3,2
*Sunday, March 31, 2013 at 4:39pm by PsyDAG*

**Stats**

If there are only two possible outcomes to an experiment then... A.) the two corresponding probabilities must each be.50. B.) the two corresponding probabilities could be any numbers between 0 and 1, but must add up to 1. C.) the two corresponding probabilities could each be ...
*Wednesday, November 2, 2011 at 9:33pm by Mandy*

**statistics**

Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores. Either-or probabilities are found by adding the individual probabilities. For last Q, ...
*Thursday, April 25, 2013 at 3:28pm by PsyDAG*

**Math**

Here are the possibilities: HHT HTH THH Each one has a probability of (1/3)(1/3)(2/3) = 2/27. For either-or probabilities, add the probabilities of the individual events.
*Monday, May 3, 2010 at 11:06pm by PsyDAG*

**Help with Calculating Probabilities**

Anon: Do you know how to calculate probabilities like P10, P20, P30...? I do not understand lot of this statistics stuff and desperately need help. Thanks
*Thursday, September 13, 2012 at 10:12pm by Bev*

**probability and statistics. **

I think there's a mistake with the probabilities were given. 1) probabilities can't be >1 (you get greater than 100% chance). 2) the probabilities don't add up to 1 here is what I think they should be: .32, .12, .23, .18, .15 P(x>3.5) = probability that a customer ...
*Tuesday, May 17, 2011 at 9:52pm by TutorCat*

**statistical analysis**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. a. Without replacement: 6/8 * 5/7 * 4/6 = ? b. You want probability of one or both being chosen. Either-or probabilities are ...
*Thursday, August 1, 2013 at 12:52am by PsyDAG*

**Statistics**

For either-or probabilities, you add the individual probabilities. .023 + .023 = ? For winning on both/all you multiply the probabilities. .023 * .023 = ? The first answer is for winning on either one of the plays, while the second is for winning on both ("or more"). From the ...
*Monday, October 29, 2012 at 8:47pm by PsyDAG *

**Math**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 1. $10 and $20 are even. The chances at each drawing are 1/2. 2. $1 and $5 are odd. The chances at each drawing are 1/2. 3. Same...
*Friday, February 8, 2013 at 2:15pm by PsyDAG*

**MATH**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. a) (1-.05)(1-.21) = ? b) .05(1-.21) = ? c) Follow same process. d) This is saying that either h or k will qualify. Either-or ...
*Thursday, November 7, 2013 at 10:38am by PsyDAG*

**problem stats**

For either-or probability, you add the probabilities of the individual events. Since all the probabilities are the same, you can just multiply the probaiblity by 5. I hope this helps. Thanks for asking.
*Monday, October 15, 2007 at 8:29pm by PsyDAG*

**math**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. a) .3^2 * .7^3 = ? b) Same as 1, 2 or 3 people agreeing. for one: .3 * .7^4 = ? for two: see above for three: .3^3 * .7^2 = ? ...
*Thursday, June 20, 2013 at 11:42am by PsyDAG*

**Statistics**

"At least one" = either 1, 2 or 3. If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. for 1: .0479 * (1-.0479)^2 = ? for 2: .0479^2 * (1-.0479) = ? for 3: .0479^3 = ? Either-or ...
*Thursday, June 13, 2013 at 8:46pm by PsyDAG*

**Statistic**

K = .40 * .40 = ? L = .30 *.60 = ? Either-or probabilities are found by adding the individual probabilities.
*Tuesday, March 18, 2014 at 1:38pm by PsyDAG*

**Math**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. "At least one" = one or more. 1/4 * (3/4)^4 = ? (one correct) (1/4)^2 * (3/4)^3 = ? (2 correct) (1/4)^3 * (3/4)^2 = ? (3 correct...
*Monday, April 29, 2013 at 1:09pm by PsyDAG*

**math**

The probability of both/all events occurring is determined by multiplying the probabilities of the individual events. a) 26/52 * (26-1)/(52-1) = ? b) Either-or probabilities are found by adding the individual probabilities. The probability above for red plus the same ...
*Monday, November 26, 2012 at 10:02pm by PsyDAG*

**math**

With 2 die, there are 36 possibilities. Each possibility has a 1/36 chance of occurring. How many ways can you get 6? 5,1; 1,5; 4,2; 2,4; 3,3. Either-or probabilities are found by adding the individual probabilities.
*Tuesday, November 27, 2012 at 11:24am by PsyDAG*

**Math**

P(4) = 1/12 P(5) = 1/12 Either-or probabilities are found by adding the individual probabilities.
*Sunday, December 15, 2013 at 6:24pm by PsyDAG*

**stats**

You need to assume that the age distribution is normal. Again, normalize the data which have a distribution of ~N(66,4²) and compute probabilities based on the looked up probabilities.
*Saturday, August 4, 2012 at 10:01am by MathMate*

**statistics**

You are not answering the questions that are being asked. They are asking for probabilities. We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though...
*Friday, June 7, 2013 at 8:35am by PsyDAG*

**math**

a. Cannot draw diagram here. Probability of each color = 4/12 = 1/3 b. First sock = 4/12, second sock = 3/11. Probability of events all occurring is found by multiplying individual probabilities. However, you want this for either blue, white or gray. Either-or probabilities ...
*Sunday, August 15, 2010 at 4:23pm by PsyDAG*

**statistics**

A professional gambler claims that he has loaded a die so that the outcomes of 1,2,3,4,5,6 have corrresponding probabilities of 0.1, 0.2, 0.3, 0.4, 0.5, and 0.6. Can he actually do what he has claimed? Is a probablility distrubion described by listening the outcomes along with...
*Sunday, May 29, 2011 at 11:51pm by Jessica*

**math asap**

"Greater than 4" means either a 5 or 6. The probability of either is 1/6. Either-or probabilities are found by adding the individual probabilities. 1/6 + 1/6 = ?
*Wednesday, December 12, 2012 at 10:49pm by PsyDAG*

**math**

Add the probabilities of getting I assume you want the probaility of getting 2 to 9 INCLUSIVE. You could add the probabilities of getting 2,3,4,5,6,7,8 and 9 OR, subtract from 1, the probabilities of getting 10,11, or 12. Let's do it the latter way, since it's quicker. 1 - 3/...
*Wednesday, August 25, 2010 at 11:42pm by drwls*

**Math**

This is the same as both are either red or green. red = (20-5-9)/20 green = 5/20 Either-or probabilities are found by adding the individual probabilities. This is for one choice. But you are looking fro two choices. If the events are independent, the probability of both/all ...
*Saturday, November 2, 2013 at 9:36am by PsyDAG*

**math probablity- Respond as soon as possible **

Assume a family is planning to have three children. 1. Why do probabilities centered around this scenario represent the same probabilities as those for flipping three coins
*Friday, April 26, 2013 at 10:28pm by Angela*

**math**

A single card is selected from an ordinary deck of cards. The sample space is shown in the figure below. Find the probabilities. (Enter the probabilities as fractions.) (a) P(two of diamonds) 1 (b) P(two) 2 (c) P(diamond) 3
*Friday, June 7, 2013 at 12:24am by Jean Claude*

**Statistics Help**

From your data, you have 6 candidates, with 2 having 9+ years of experience and 3 being female. This should give you the probabilities of A and B. Assuming you want to know what is the probability that the candidate will be both experienced and female, multiply the two ...
*Sunday, March 29, 2009 at 4:50pm by PsyDAG*

**probability**

Calculate the numerical values of following probabilities, as well as the expected value and variance of X: P(X=0)= P(X=1)= P(X=−2)= P(X=3)= E[X]= var(X)= Let Y=X2. Calculate the following probabilities: P(Y=0)= P(Y=1)= P(Y=2)=
*Thursday, February 27, 2014 at 5:51pm by Anonymous*

**algebra**

Getting "a 5 at least once" = one, two or three 5s. Once = 1/6(5/6)(5/6) = ? Twice = 1/6(1/6)(5/6) = ? Three times = 1/6^3 = ? Either-or probabilities are found by adding individual probabilities.
*Thursday, May 5, 2011 at 8:03pm by PsyDAG*

**Statistics**

Either-or probabilities are found by adding the individual probabilities. Since three face cards are already spades, delete them for the face card probability. 9/52 + 13/52 = ?
*Saturday, February 22, 2014 at 4:08pm by PsyDAG*

**stat**

The probability of answering 5/6 = (1/4)^5 * 3/4 The probability of answering all 6 = (1/4)^6 Either-or probabilities are found by adding the separate probabilities.
*Sunday, June 13, 2010 at 6:35pm by PsyDAG*

**Math**

That's 25 school days. The average or "expectation" number of late appearances will be 25x0.04 = 1.0, but there are non-zero probabilities of 0,2,3, etc. late appearances as well. The binomial distribution can be used to calculate their probabilities.
*Sunday, May 23, 2010 at 11:01am by drwls*

**statistics tim**

Assuming that the percentage applies to an hour, First item p=.05 Second item p = .05 The probability of all/both probabilities occurring is found by multiplying the individual probabilities.
*Monday, February 28, 2011 at 9:08pm by PsyDAG*

**Math**

We cannot show a tree diagram via this medium. However, it would be easier to start your tree with A and B. Chances of 1 = 1/6 Chances of A = 1/2 To find the probability of both (or all) probabilities occurring, multiply the individual probabilities.
*Friday, April 23, 2010 at 12:36am by PsyDAG*

**MATH Prob.**

You need to use the two probabilities, 7/17 and (since one red has been removed) 6/16. The probability of both occurring is found by multiplying the probabilities of the individual events. Try it, and one answer will pop out for you.
*Saturday, August 8, 2009 at 1:55pm by PsyDAG*

**statistics**

1. .65 * (1-.12) = ? 2, 3. Right! 4. .07 * .35 = ? Either-or probabilities are found by adding the individual probabilities. 5. .65 + (.88*.65)(.93 *.35) = ?
*Thursday, April 18, 2013 at 11:40am by PsyDAG*

**Stats**

Probability of picking pepperoni on the first pick = 1/13. Assuming that customer does not want to "double up" on any topping, 12/13 for first pick for some other topping and 1/12 for second for pepperoni. If the events are independent, the probability of both/all events ...
*Wednesday, July 17, 2013 at 9:42pm by PsyDAG*

**Statistics**

Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to those Z scores. For either-or probabilities, add the individual probabilities.
*Saturday, October 8, 2011 at 11:43pm by PsyDAG*

**Math Probability**

I would assume that you determined the two probabilities separately. Red cards = 26/52 5 = 4/52 Either-or probability determined by adding the individual probabilities.
*Tuesday, April 27, 2010 at 12:12am by PsyDAG*

**Math**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 8/15 * 7/15 = ? Either-or probabilities are found by adding the individual probabilities. 4/41 + 10/41 = ? With replacement, 6 ...
*Friday, January 11, 2013 at 6:51pm by PsyDAG*

**Statistics**

A. If it is the first pick, it doesn't matter what the subsequent picks are, 10/30. B. More than two women means 3, 4 or 5 women. If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events...
*Saturday, May 18, 2013 at 10:24am by PsyDAG*

**statistics**

The easiest way to do this problem is to use a binomial distribution table. For a), n = 20, p = .4, x = 0, 1, 2 Add together the probabilities you find in the table, then subtract from 1 for your answer. For b), n = 20, p = .4, x = 0, 1, 2, 3, 4, 5 Add together the ...
*Friday, March 16, 2012 at 1:57pm by MathGuru*

**math 157**

There are 13 spades and 13 hearts in the 52-card deck (13/52 and 13/52, respectively). For "either-or" probabilities, you add the probabilities of the individual events. I hope this helps.
*Sunday, January 17, 2010 at 6:38pm by PsyDAG*

**statistics**

The probability of all events occurring is found by multiplying the probabilities of the individual events. .7^6 * .3^4 = ? (exactly six) Add probabilities of 6, 7, 8, 9 and 10 for "at least." e.g., for 10 wearing = .7^10 = ? for 9 wearing = .7^9 * .3 =? I'll let you do the ...
*Thursday, June 17, 2010 at 4:36pm by PsyDAG*

**statistics**

31 + 13 + 3 ≠ 50 What about the remaining 3 students? 13/50 = business major 31/50 = Democrats Either-or probabilities are found by adding probabilities of individual events.
*Sunday, August 8, 2010 at 6:05pm by PsyDAG*

**math**

This is the same as finding one or two face cards. If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. Two face cards = 12/52 * 11/51 = ? One face card = 12/52 * (51-11)/51 = ? ...
*Sunday, March 3, 2013 at 11:47pm by PsyDAG*

**math**

Again, either-or probabilities are found by adding the individual probabilities. Each combination of the die has a 1/36 probability. a) What are the possible number of doubles? Follow the rule above. b) This is the same as 1-probability of getting 2 or less with two die. c) ...
*Monday, November 26, 2012 at 10:03pm by PsyDAG*

**Statistics**

Suppose you conduct a study and find that the probability of having a baby boy is 60%. Now suppose three of your relatives are going to have babies. a) Build a tree diagram showing all the conditional probabilities and joint probabilities associated with the sex of the three ...
*Saturday, May 10, 2008 at 11:24pm by Danny*

**Statistics**

Suppose you conduct a study and find that the probability of having a baby boy is 60%. Now suppose three of your relatives are going to have babies. a) Build a tree diagram showing all the conditional probabilities and joint probabilities associated with the sex of the three ...
*Sunday, May 11, 2008 at 10:14am by Danny*

**math**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 11/15 * 11/15 = ? That means one or both do not work. One not work = 4/15 * 11/15 = ? Both not work = 4/15 * 11/15 = ? Either-or...
*Thursday, June 6, 2013 at 1:37pm by PsyDAG*

**Statistics-Matrix of Transition Probabilities**

Given the following matrix of transition probabilities, write three equations that, when solved, will give the equilibrium state values. P = large bracket with a b on top and c d directly under that, closed large bracket. Thanks.
*Monday, February 6, 2012 at 8:13am by Lisa*

**Finite Math-Probability**

Assuming that the balls are replaced, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. a) (6/16)^3 * (10/16)^3 = ? If they are not replaced, a) 6/16 * 5/15 * 4/14 * 10/13 * 9/12 * 8/11 = ? Depending on ...
*Saturday, November 24, 2012 at 5:05pm by PsyDAG*

**Statistics**

1) If I am looking at this correctly, I don't see any number that starts with 2, but I see one number that ends with 5 You probability is .079 2) I don't see any number beginning with 4. The probability that it won't be a 4 is the sum of all of the probabilities in your list ...
*Sunday, March 10, 2013 at 11:54pm by Dr. Jane*

**stats.**

a-c. Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores. d. If the events are independent, the probability of both/all events occurring is ...
*Sunday, March 17, 2013 at 9:56pm by PsyDAG*

**statistic****help*******

We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though. However, I will start you out. 10. Since you are dealing with percents, the grand total = 1....
*Wednesday, March 19, 2014 at 8:20am by PsyDAG*

**math**

The numerals get smaller if you are considering probabilities without replacement. In other words, using a numeral in one position would prohibit its use in other positions. However, the use of any numeral does not effect the probabilities of the remaining numerals. I hope ...
*Tuesday, February 26, 2008 at 3:55pm by PsyDAG*

**statistics**

μ=98.2 σ=0.62 Z(100)=(100-98.2)/0.62=2.903 Look up (normal) probabilities for Z=2.903 from tables (for people with temperatures below 100). Subtract from 1 to get probabilities of healthy person considered having a fever.
*Sunday, May 19, 2013 at 7:00am by MathMate*

**statistics**

Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. P(even) = 10/21 Multiples of 3 are 3, 6, 9, 12, 15, 18, 21. Probability of either/or = sum of individual probabilities - common probabilities.
*Sunday, December 4, 2011 at 3:53pm by PsyDAG*

**Math**

Probability that both/all events would occur is found by multiplying the individual probabilities. .86 * .028 = ? Either-or probability is found by adding the individual probabilities. (.86 * .028) + (.12 * .044) + (.02 * .035) = ?
*Sunday, October 3, 2010 at 4:45pm by PsyDAG*

**harvard**

6. For Design 3, consider the probability of a random individual in South Dorchester being sampled; and the probability of a random individual in Harbor Islands being sampled. These probabilities are approximately the same as the probabilities calculated using:
*Saturday, December 8, 2012 at 4:05am by jasonsie*

**math**

you roll a number cube. what are the odds in favor of rolling a 1 or a 5?thanks. There are 6 sides of a cube. You want one of 2 numbers to show up. So it's 2/6 chance....or 1/3 Matt I assume that you are talking about a die (Singular for "dice"), it has a total of 6 sides. ...
*Friday, April 13, 2007 at 1:16pm by dillon*

**Statistics**

Probability of all/both events occurring is found by multiplying the individual probabilities. A) 1/5 * 1/2 = ? Probability of either/or events occurring is found by adding the individual probabilities. B) Follow above instructions. c) (1-1/5)(1-1/2) = ? This info should lead ...
*Friday, July 8, 2011 at 2:47pm by PsyDAG*

**Statistics**

We don ot have the data that is "shown here." However, for "West and No," find that cell and divide by the grand total. For the "South or Midwest or Yes," find each of those totals and divide each by the grand total. For "either-or" probabilities, you have to add these ...
*Monday, July 20, 2009 at 4:51pm by PsyDAG*

**Math**

a. = total female/grand total b. = (total finance/grand total) + (total accounting/grand total) When you want either-or probabilities, you need to add the individual probabilities. e. This sounds like you want the male accounting majors out of the total males. I hope this ...
*Wednesday, June 11, 2008 at 9:51pm by PsyDAG*

**Stats (probability)**

(i) Choosing the first girl with blue eyes = 3/10. Since there is no replacement, the probability of the second girl having blue eyes = 2/9. The probability of both/all occurring is found by multiplying the individual probabilities. (ii) Do similar process with non-blue ...
*Wednesday, March 16, 2011 at 12:17am by PsyDAG*

**Math**

Either-or probability is found by adding the individual probabilities. When you are concerned with the probability that all/both the events would occur, you multiply the individual probabilities. I will demonstrate with b. Probability of getting a female bullfrog is .3 * .4...
*Monday, July 7, 2008 at 1:36am by PsyDAG*

Pages: **1** | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | Next>>