Search: math..probabilities

Number of results: 188,494

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

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

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

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
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

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

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

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
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
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

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

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

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

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 ...
Monday, May 13, 2013 at 1:08pm 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
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

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

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
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
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...
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 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
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

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

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

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

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
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
< 5cups = (1 - .24) Either-or probabilities are found by adding the individual probabilities.
Monday, May 6, 2013 at 8:00pm 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...
Wednesday, August 25, 2010 at 11:42pm by drwls

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
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

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

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
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 ...
Thursday, April 25, 2013 at 3:28pm 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 ...
Tuesday, May 17, 2011 at 9:52pm by TutorCat

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"...
Monday, October 29, 2012 at 8:47pm 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

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

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 = ? ...
Thursday, June 13, 2013 at 8:46pm by PsyDAG

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

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

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

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
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 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
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

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

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

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

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 ...
Friday, April 13, 2007 at 1:16pm by dillon

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

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

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

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

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
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

Math
Without replacement, the probability of drawing the nickel is 3/10, the dime is 2/9 and the quarter is 5/8. Mulitply the probability of the three events to get the probability that all would occur. Each probability is dependent on the previous event. With replacement, the same...
Thursday, December 13, 2007 at 8:45am 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

Math
There are 36 possibilities with two dice. Either-or probabilities are found by adding the individual probabilities. 1. A sum of seven can be gotten with 2,5; 5,2; 3,4; 4,3; 6,1; or 1,6. That is 6/36 = 1/6 2. Sum greater than three = 1 - sum of 3 or less. Go through the same ...
Saturday, November 24, 2012 at 3:42pm by PsyDAG

Math
i) It is the proportion below for 2 the same, plus 1/7*1/7*1/7 for 3 the same. (Either-or probabilities are determined by adding the separate probabilities.) ii) It is 1/7*1/7*6/7, because the third person has a different day of the week. iii) Work the same as i) above, for 2...
Monday, February 22, 2010 at 2:50am 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

Elementary math College
To get the first red chip = 7/16, for the second 6/15. The probability of both occurring is found by multiplying the individual probabilities. To get the first blue chip = 9/16, for the second 8/15. The probability of both occurring is found by multiplying the individual ...
Tuesday, March 30, 2010 at 1:33pm by PsyDAG

Math Probability
A binomial distribution applies when 1. Trials have exactly two possible outcomes 2. probabilities remain constant throughout trials. 3. there is a defined number of trials. 4. trials are independent of each other. 5. the random variable is the number of successes. Since the ...
Monday, April 16, 2012 at 2:03pm by MathMate

Math
Hints: 1. If the probability of the hockey team winning is 0.65, then the probability of the team losing is 1-0.65=0.35, by Kolmogorov's third axiom. 2. The multiplication rule: The probability of two independent events happening is the product of the individual ...
Thursday, March 8, 2012 at 5:34pm by MathMate

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 ...
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

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

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...
Thursday, June 17, 2010 at 4:36pm 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 ...
Sunday, May 11, 2008 at 10:14am by Danny

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

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...
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 ...
Sunday, March 17, 2013 at 9:56pm by PsyDAG

Math
A. 4746/(grand total) = ? Grand total is the sum of all the observations. B. Either-or probabilities are found by adding the individual probabilities. (All age 20/grand total) + (all female/grand total) = ? C. (all 20+)/(grand total) = ?
Wednesday, April 10, 2013 at 8:12pm 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

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

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

math
Find the requested probabilities. P(B) = 1/5
Tuesday, September 11, 2012 at 9:30am by Anonymous

MATH: PROBABILITIES
xcvb cv
Friday, April 23, 2010 at 4:11pm by Anonymous

MATH: PROBABILITIES
any one know the answer:(?
Saturday, April 24, 2010 at 11:22pm by Anonymous

MATH: PROBABILITIES
both of you are wrong...
Saturday, April 24, 2010 at 11:22pm by Anonymous

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" ...
Monday, July 20, 2009 at 4:51pm 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

Statistics
A roster contains the names of 12 students from the School of Math, Science and Engineering (MSE) and 8 students from the Dreeben School of Education (DSE). Determine the probabilities of randomly selecting the following if the selection is done without replacement: A) An MSE ...
Monday, March 7, 2011 at 9:11pm by James

Math
Shouldn't the probabilities add to 1.0? What other possibilities are there/
Sunday, August 15, 2010 at 12:30pm by bobpursley

math: probabilities
6x5x4(Because you want them to be all different)/6x6x6
Wednesday, June 2, 2010 at 5:29pm by Jen

Mathematics
Please help... In a 200 m swimming race between 8 boys, the probability of Omar winning the race if he is in one of the two outside lanes is 1/4. If he is in any of the other lanes, the probability of Omar winning is 1/3. If the lanes are drawn at random, what is the overall ...
Tuesday, May 29, 2007 at 1:37pm by Elizabeth

probability
Binomial expansion is typically finding the coefficients of the expansion of (a+b)^n. In this particular case, a+b=1, so 1^n is still n. The individual terms therefore represent the respective probabilities of the combination of outcomes, 2R+3B, 0R+5B, etc. The beauty of this ...
Friday, June 10, 2011 at 1:50pm by MathMate

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