Saturday

April 19, 2014

April 19, 2014

Number of results: 3,120

**Math**

prob(hit) = 27/54 = 1/2 prob( miss) = 1/2 prob (no points) = prob(miss, miss) = (1/2)(1/2) = 1/4 prob( 1 point) = prob(HM) + Prob(MH) = 2(1/4) = 1/2 prob(2 points) ]= prob (HH) = 1/4 (notice 1/4 + 1/2 + 1/4 = 1 , as expected)
*Thursday, May 16, 2013 at 9:30pm by Reiny*

**Binomial Probability**

Prob(male) = .7 prob(female) = .3 a) prob(6 are male_ = C(12,6) (.7)^6 (.3)^6 = .07925 b) prob(6 or more are female) = prob(6 F) + prob(7 F) + prob(8 F + .. prob(12 F_ = C(12,6)(.3)^6 (.7)^6 + C(12,7) (.3)^7 (.7)^5 + ... + C(12,12) (.3)^12 (.7)^0 I will let you do the button ...
*Saturday, November 10, 2012 at 4:32pm by Reiny*

**statistics**

prob of R(ight) = 4/20 = 1/5 prob of W(rong) = 4/5 1) to have exactly 9R and 11W = C(20,9)((1/5)^9(4/5)^11 = .... ( I got .00722) 2) to get the prob of less than 9 you will have to do prob(0R) + prob(1R) + prob(2R) + .. + prob(8R) I will do prob(R4) = C(20,4)(1/4)^4(4/5)^16...
*Tuesday, March 23, 2010 at 2:21pm by Reiny*

**stats**

a test consists of 25 multiple choice questions. Each has 5 possible answers, which only one is correct. If a student guesses on each question, find the following: a) prob that he will guess all of them right b)prob that he will guess at most 12 right c) prob that he will ...
*Friday, February 18, 2011 at 10:09pm by gio*

**Binomial Math**

prob of purchase = .4 prob of no purchase = .6 prob of at least 5 from 10 will make purchase = prob(5will buy) + prob(6 will buy) + ..+ prob(10 will buy) .... lots of arithmetic, I will do the prob 6 will buy = C(10,6) (.4)^6 (.6)^4 = .. What might be a shorter way is to ...
*Sunday, January 27, 2013 at 2:48pm by Reiny*

**stats**

prob that a new Canadian can swim = .82 let that be Y prob that a new Canadian cannot swim = .18 let that be N so we are looking at something like NNNNNNYYY that particular prob would be (.18^6(.82)^3 but the NNNNNNYYY can be arranged in 9!/(6!3!) ways or C(9,6) or 84 ways so ...
*Saturday, October 2, 2010 at 11:40am by Reiny*

**AP Stats**

make a "tree" of possiblilites start with two branches A) took SAT prep test B) did not take SAT prep test mark each with a prob of .5 Split A) into two more braches C) got in to first choice college, prob .3 D) did not get into first choice college, prob .7 Split B) into two ...
*Friday, January 2, 2009 at 2:39pm by Reiny*

**MTH 156**

Prob(AorBorC)=Prob(A)+Prob(B)+ Prob(C)-Prob(AandB)-Prob(BandC)-Prob(AandC) + Prob (AandBandC). so. cranking it out...check closely... Prob(AorBorC)=2/11+ 3/11+ 4/11 - 2/11*3/11 - 3/11 * 4/11 - 4/11*2/22 + 0 81/121 -24/121= 57/121 check that.
*Friday, September 11, 2009 at 6:02pm by bobpursley*

**Math**

prob of a 6 = 1/6 so prob(three consecutive 6's) = (1/6)(1/6)(1/6) = 1/216 b) prob(any particular number) = 1/6 so prob(5,1,then even) = (1/6)(1/6)(1/2) = 1/72 c) prob(odd, >2, 5) = (1/2)(4/6)(1/6) = 4/216 = 1/54
*Wednesday, April 10, 2013 at 10:56am by Reiny*

**prob and stats**

0-9-
*Friday, October 15, 2010 at 1:14am by 0-0--*

**Stats**

One possible outcome is GGRRR the prob of that is (8/18)(7/17)(10/16)(9/15)(8/14) = 2/51 But the 2 G's and the 3 R's can be arranged in 5!/(2!3!) or 10 ways so prob of your event = 10(2/51) = 20/51
*Monday, June 6, 2011 at 4:30pm by Reiny*

**Stats**

prob of tail --- x prob of head --- 2x x+2x = 1 3x=1 x = 1/3 expected win = (1/3)7 + (2/3)(-3) = 1/3 or $.33 but you are paying $5.00 for a return of .33, silly! "The expected valve of net gain is positive and the player should not play "
*Friday, April 12, 2013 at 4:39pm by Reiny*

**Algebra 2: Prob and Stats**

Thank you so much!
*Monday, May 24, 2010 at 7:06pm by Skye*

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

mmmh, in flipping one coin, and having one child: prob(heads) = 1/2, prob(tails) = 1/2 prob(boy) = 1/2 , prob (girl) = 1/2 for flipping 3 coins, or considering 3 kids: prob(1 head, 2 tails) = 3(1/2)^3 = 3/8 prob(1 boy, 2 girls) = 3(1/2)^3 = 3/8 etc.
*Friday, April 26, 2013 at 10:28pm by Reiny*

**Algebra**

Prob(A U B) = Prob(A) + Prob(B) - Prob(A ∩ B) I got .88 or 22/25
*Tuesday, March 11, 2008 at 3:36pm by Reiny*

**college**

Prob(A OR B) = Prob(A) + Prob(B) - Prob(A AND B) = .5 + .65 - .18 = .97 looks like C)
*Monday, April 13, 2009 at 12:56pm by Reiny*

**stats & prob.**

Steve you really said that? :(
*Tuesday, November 27, 2012 at 3:21pm by Anonymous*

**stats & prob.**

Stop cheating! Answer your own questions.
*Tuesday, November 27, 2012 at 3:21pm by Steve*

**stats & prob.**

Stop cheating! Answer your own questions.
*Tuesday, November 27, 2012 at 3:21pm by Steve*

**math**

a) 3 outcomes: GG , GB, BB b) 3 G, and 4B prob (GG) = (3/7)(2/6) = 1/7 c) prob (BB) = ((4/7)(3/6) = 2/7 d) prob(not both blue or grey) = prob (BG) + prob(GB) = (4/7)(3/6) + (3/7)(4/6) = 2/7 + 2/7 = 4/7 or we could have taken 1 - (1/7+2/7) = 4/7
*Thursday, November 29, 2012 at 7:30pm by Reiny*

**MATH**

You have a probability of getting a penny is 25/50 prob nickel is 10/50 prob of dime is 10/50 prob quarter is 5/50 Prob(x<13)=prob(penny)+prob(dime)+prob(nickel)= 45/50
*Monday, November 2, 2009 at 4:05pm by bobpursley*

**Stats**

prob(PPP) = (10/23)(9/22)(8/21) = 120/1771
*Tuesday, July 10, 2012 at 6:34pm by Reiny*

**Maths**

Say the probability he'll pass English is Prob(E) = 0.6. The probability he'll pass in both English and Maths is Prob(E&M) = 0.54. Provided the probability that he'll pass English is independent of the probability that he'll pass Maths (and note that's an assumption we're ...
*Monday, May 2, 2011 at 12:56pm by David*

**stats & prob.**

Given the values of n=5, p= 0.15, and (1-p)=0.85, calculate the standard deviation of the data.
*Tuesday, November 27, 2012 at 3:21pm by Georgia*

**Prob and stats\ math**

How many Learners who are passed this year at Hans kekana high school?
*Wednesday, December 20, 2006 at 8:44am by Innocent Mathulatshipi*

**Maths**

let prob of success be p let prob of failure be q, where p+q = 1 Prob(3out of 5) = C(5,3) p^3 q^2 = 10p^3 q^2 prob(2 out of 5) = C(5,2)p^2 q^3 = 10 p^2 q^3 10p^3q^2/(10p^2q^3 = 1/4 q = 4p in p+q=1 p + 4p=1 p = .2 , then p = .8 so prob(4 out of 6) = C(6,4) (.8)^4 (.2)^2 = ....
*Tuesday, May 15, 2012 at 9:40am by Reiny*

**Probability**

Prob(5plain tables) on Monday = .8^5 prob(5 plain) on Tuesday = .8^5 prob(5 plain on Monday AND 5 plain on Tuesday) = (.8^5)(.8^5) = .8^10 = .107 Prob(at least one deluxe) = 1 - Prob(all 5 plain) = 1 - .8^5 = .672 b. don't know about Poisson random variables.
*Wednesday, January 21, 2009 at 11:49am by Reiny*

**Probability and statistics**

Prob(2) = 1/36 prob(3) = 2/36 prob(4) = 3/36 prob(12) = 1/36 prob(11) = 2/36 prob(10) = 3/36 total of above , your cases of winning = 12/36 so the prob of remaining cases = 24/36 expected value of game = (12/36)(5) + (24/36)(-5) = (1/3)(5) - (2/3)(5 = -5/3 You would be ...
*Tuesday, September 18, 2012 at 4:25pm by Reiny*

**probability**

prob of defect = .1 prob of NOT defect = .9 4 or more defective means exclude cases of 0, 1, 2, or 3 defective prob of none defective = C(15,0) (.1^0)( .9^15) = .20589 prob of one defective = C(15,1) (.1)^1 (.9)^14 =.34315 prob of two defective = C(15,2) (.1^2)(.9^13) = ....
*Sunday, July 22, 2012 at 10:48pm by Reiny*

**math**

Prob of 1 or 2 on a die = 2/6 = 1/3 prob of 2 or 4 = 1/3 prob of 5 or 6 = 1/3 Since these become your choice of answer prob of selecting 1st answer = 1/3 prob of selecting 2nd answer = 1/3 ..... so the prob of choosing the correct answer in each event is simply 1/3 so to get 7...
*Tuesday, November 20, 2012 at 6:48am by Reiny*

**Math**

prob of A = .15 prob of notA = .85 a) prob all 3 to get A = (.15)^3 = .003375 b) exactly 2 A's = C(3,2) .15^2 (.85) = .057375 c) at least one = 1 - prob no A's = 1 - .85^3 = .385875 or the long way: prob oneA + prob 2 A's + prob 3 A'a = C(3,1) (.15)(.85)^2 + C(3,2) (15)^2 (.85...
*Tuesday, August 27, 2013 at 4:26am by Reiny*

**probability**

This is a case of binomial distribution Prob(watching) = .68/100 = 17/25 prob(not watching) = 8/25 a) Prob(exactly 6 out of 12 watching) =C(12,6)(17/25)^6(8/25)^6 = .... b) prob(6 or less) = Prob(exactly 1) + Prob(exacly 2) + prob(exactly 3) + .. prob(exactly 6) = C(12,1)(17/...
*Thursday, October 14, 2010 at 1:11am by Reiny*

**math probability**

the first 3 are correct for d) the odds are 1,3,5,7,9,11,13 multiples of 3 are 3,6,9,12 now the numbers which are either odd OR a multiple of 3 are 1,3,5,6,7,9,11,12,13 or nine of them so prob = 9/13 There is a formula which says Prob(A or B) = Prob(A) + Prob(B) - Prob(A and B...
*Thursday, December 18, 2008 at 12:22am by Reiny*

**Probability**

let N be a normal coin, and let DH be a double - headed coin let DT be double-tailed H -- heads, T --- tails for N, prob(H) = 1/2, prob(T) = 1/2 for DH, prob(H down) = 1, prob(Tdown) = 0 for DT, prob(H down) = 0, prob(Tdown) = 1 so you could draw N or DH or DT prob(heads down...
*Friday, March 7, 2014 at 2:31pm by Reiny*

**algebra**

prob(2) = 1/6 prob(not 2) = 5/6 prob (four 2's out of 6) = C(6,4) (1/6)^4 (5/6)^2 = 15(1/1296)(25/36) = appr .008
*Tuesday, December 3, 2013 at 2:10pm by Reiny*

**probability**

Prob(B or C) = Prob(B) + Prob(C) - Prob(B and C) = .35+.63 - .4 = ...
*Wednesday, February 15, 2012 at 8:43am by Reiny*

**Math**

a) So it could be GBB or BGB or BBG prob of that is 3(1/2)(1/2)(1/2) = 3/8 b) at most 2 boys ---> cannot have BBB which has a prob of 1/8 so your case has prob of 1 - 1/8 = 7/8
*Monday, January 10, 2011 at 3:22pm by Reiny*

**Algebra 2: Prob and Stats**

Find the sample size that produces the margin of error +-4.0% a) 325 b) 25 c) 16 d) 625
*Monday, May 24, 2010 at 7:07pm by Skye*

**math**

a) prob = 47/69 b) prob = 20/51 c) prob = 67/120 Unless I am missing something, this looks pretty straightforward.
*Tuesday, March 30, 2010 at 9:39pm by Reiny*

**prob and stats (incomplete)**

A and C are not complete. Copy and paste may not work here. B. mean = 72.15 SEm = SD/√n
*Wednesday, October 24, 2012 at 3:29pm by PsyDAG*

**Probability**

I will assume that you filled in the Venn diagrams correctly n(M upside down u W) ---> n(M and W) = 14 n(M' U S) ---> n(M' or S) Since S is the symbol used for the universal set, the count would be 49 P( both mice are short-tailed) --- where does the "both" come from...
*Thursday, February 20, 2014 at 8:30pm by Reiny*

**Finite Math**

To get 40% or more, you cannot have 0, 1, 2, or 3 only correct answers prob(0 right) = C(10,0) (1/5)^0 (4/5)^10 = .10737 prob(1 right) = C(10,1)(1/5)(4/5)^9 = .26844 prob(2right) = C(10,2)(1/5)^2(4/5)^8 = .30199 prob(3right) = C(10,3)(1/5)3 (4/5)^7 = . 20133 total = .87913 so ...
*Monday, May 23, 2011 at 8:43pm by Reiny*

**math30**

Prob(ball) = 2/4 = 1/2 prob(parachute) = 1/4 prob(frisbee) = 1/4 prob(chicken) = 1/3 prob(fish) = 1/3 prob( chicken or fish AND a ball) = (1/3)(1/2) + (1/3)(1/2) = 1/6 + 1/6 = 1/3
*Friday, January 25, 2013 at 3:28am by Reiny*

**math-probaility**

There are only 3 possibilites: - no lemon - 1 lemon - 2 lemon so you want the prob (1 lemon OR 2 lemon) = 2(3/5)(2/4) = 3/5 + (3/5)(2/4) = 3/10 = 9/10 This involved finding the prob of two cases, plus an addition. You know that the prob of all 3 cases is 1 so what MathMate did...
*Saturday, June 1, 2013 at 10:16am by Reiny*

**Math**

Prob of at least B is = prob of A +prob of B= (45+180)/totalofallgrades
*Wednesday, July 20, 2011 at 10:08am by bobpursley*

**Probabilities**

This conditional probability the formula is P(A│B), read the prob of A given B = P(A and B)/P(B) in your case Prob(A) is Prob(green) B is "not red or blue" so find Prob(green AND "not red or blue") and Prob(not red or blue) and sub into the formula
*Tuesday, April 21, 2009 at 2:13pm by Reiny*

**math**

generally you don't subtract 1, you subtract FROM 1 The prob. of anything is a number between 0 and 1, so often when there are many cases to consider, it might be easier to calculate the prob of the exceptions, then subtract that from 1. e.g. What is the probability of ...
*Thursday, August 26, 2010 at 3:59pm by Anonymous*

**Math**

prob of heading bull's eye = 15/20 = 3/4 So the prob of not heading it is 1/4 odds in favour of some event = prob(of event) : prob(not the event) = (3/4) : (1:4) = 3:1
*Monday, February 8, 2010 at 10:19am by Reiny*

**math**

N(C or D) = N(C) + N(D) - N(C and D) = 40 + 25 - 15 = 50 prob(C or D) = 50/100 = 1/2 N(Cat or Dog, not both) = 40 + 25 - 15 -15 = 35 Prob(that event) = 35100 = 7/20 c) Prob( A | B) ----- conditional prob = Prob( A and B)/Prob(b) prob(dog | cat) = prob(dog and cat)/prob(cat...
*Sunday, February 17, 2013 at 11:41am by Reiny*

**Stats**

RANDOM can be arranged in 6! or 720 ways. prob of getting it in that form = 1/720
*Tuesday, April 15, 2014 at 7:14am by Reiny*

**Statistics**

For any given birth-month the guesser wins by guessing the 5 months "around" that month. e.g. if born in June, the correct choices would be April, May, June, July , and August, which is 5 months. Prob(guesser wins) = 5/12 prob(1 win out of 6) = C(6,1) (5/12) (7/12)^5 = .... ...
*Tuesday, July 3, 2012 at 6:22am by Reiny*

**statistics - math**

I don't have your Appendix, nor do I know which Excel function you are talking about, but .. Prob (female) = 60/100 = 3/5 prob (male) = 2/5 prob (5 of 13 are female) = C(13,5)(3/5)^5 (2/5)^8 = 1287 (.07776)(.0006553) = .06559 do prob(6 of 13 are female) the same way and add up...
*Saturday, December 4, 2010 at 11:17pm by Reiny*

**Probability/Stas**

actually their answer is wrong as well you want the prob of losing 6 times in a row, so if prob of winning is .023 then the prob of losing in a game is .977 so prob of losing 6 consecutive times = (.977)^6 = .8696958
*Tuesday, March 30, 2010 at 6:46pm by Reiny*

**Prob Stats**

The probability of both/all events occurring is found by multiplying the probabilities of the individual events. .61^6 = ?
*Monday, January 9, 2012 at 7:51am by PsyDAG*

**Algebra II**

1. prob(a 3) = 1/6 prob(greater than 3) = 3/6 = 1/2 since you want prob a red 3 AND you would have (1/6)*(1/2) = 1/12 How did you get 3/5 ? 2. prob(first greater than 25) = 26/50 prob(second greater than 25) 25/49 prob (third greater than 25) = 24/48 so ... 26/50 * 25/49 * 24/...
*Sunday, March 30, 2008 at 4:42pm by Reiny*

**Math**

let's look at the prob that they are all different start by picking any glove, now you have 1 there is 1 of the remaining 9 that will match we don't want that, so the prob that the 2nd is NOT a match is 8/9 prob that the 2nd and third are NOT a match = (8/9)(7/8) prob that the...
*Tuesday, October 2, 2012 at 7:43pm by Reiny*

**Mathematicas**

there are 8 ways to get a sum of 9 1 8, 2 7, .... , 8,1 let's look at the prob of getting one of those pairs, the 1 8 prob of getting the 1 is 1/10. since you are replacing the card, the prob of getting an 8 on the second draw is also 1/10 so the prob of getting the 1 8 ...
*Thursday, January 28, 2010 at 10:53pm by Reiny*

**URGENT MATH!!!!!!!**

in #1, are you picking just one? I will assume that Prob(1 red) = 4/35 = appr .114 which is 11.4% you had the right answer #2 There are 4 numbers > 3 so prob (>3) = 4/6 = 2/3 (they should have reduced the fractions) #3 prob of correct guess = 1/5 prob of wrong guess = 4/...
*Tuesday, April 16, 2013 at 11:37pm by Reiny*

**Math**

prob(6) = 1/6 prob(not6) = 5/6 a) exactly 1 out of 7 tries to be 6 = C(7,1) * (1/6)^1 * (5/6)^6 = .3907 rounded to 4 decimals b) at least one 6 ---> 1 - prob(all not6) = 1 - C(7,7) * (1/6)^0 * (5/6)^7 = 1 - .2791 = .7209
*Tuesday, October 25, 2011 at 8:15am by Reiny*

**Math**

prob of losing = 8/100 = 2/25 prob of not losing it = 23/25 prob(2 out of 14losing it) = C(14,2) (2/25)^2 (23/25)^12 = appr .214 b) prob (at least 12) = Prob(12) + prob(13) + prob(14) = C(14,12) (2/25)^12 (23/25)^2 + .....
*Wednesday, December 8, 2010 at 12:16am by Reiny*

**math**

primes on a die are 2,3 and 5 so prob of a prime = 3/6 = 1/2 and prob NOT prime = 1-1/2= 1/2 prob of no prime = C(5,0) (1/2)^5 = 1/32 prob of one prime = C(5,1) (1/2)(1/2)^4 = 5/32 prob of at least 2 primes = 1 - 1/32 - 5/32 = 26/32 = 13/16 or .8125
*Saturday, March 17, 2012 at 9:19am by Reiny*

**Math**

in the first, the prob of getting the B is 2/13, replacing that and then picking a T has a prob of 1/13 so the prob of picking a B, followed by the T is (2/13)(1/13) = 2/169 in the second you are not replacing the letter so for the second prob. there are only 12 letters left ...
*Monday, May 25, 2009 at 9:37am by Reiny*

**math**

a) prob (exactly3) = C(7,3) (.35)^3 (.65)^4 = .2679 b) at least 3 people = 1 -(prob(none) + prob(one) + prob(two) = 1 - ( C(7,0) .65^7 + C(7,1) (.35)(.65)^6 + C(7,2)(.35)^2 (.65)^5 ) = ..... you do the button pushing. c) at most 5 ---- > 0,1,2,3,4,5 or exclude: 6 and 7 d) ...
*Wednesday, April 3, 2013 at 3:49pm by Reiny*

**Algebra 2 - stats and probability**

1. Let E be 'eagle on back" let N be 'not eagle on back' There are only 4 possible outcomes EE, NE, EN, and NN since there is replacement each of those above events has a prob of 1/4 from (2/4 x 2/4) so a) is 2/4 x 2/4 = 1/4 b) NN or 1/4 again c) is 1/4 + 1/4 = 1/2 3. for ...
*Wednesday, May 13, 2009 at 1:37pm by Reiny*

**math**

prob of liking = .9 prob of not liking = .1 prob that 2 of 5 will like = C(5,2)(.9)^2 (.1)^3) = 10(.81)(.001) = .0081
*Monday, March 12, 2012 at 5:10pm by Reiny*

**Prob and stats**

A. There is only one queen of clubs, 1/52. B. There are two red fours, 2/52. C. There are four sevens, 4/52.
*Tuesday, February 19, 2013 at 12:02am by PsyDAG*

**prob and stats**

This type of question can be answered using a table or website of the normal-distribution function. Using (Broken Link Removed) I get 95 to be the cutoff grade for an A
*Friday, October 15, 2010 at 1:13am by drwls*

**Algebra II**

not quite, 4/52 is the probability of drawing a king, which includes the 2 red kings 26/52 is the prob. of drawing a red card, which includes the two red kings, which you already accounted for so 4/52 + 26/52 - 2/52 = 28/52 = 7/13 I am using the formula: Prob(A OR B) = Prob(A...
*Sunday, March 30, 2008 at 5:21pm by Reiny*

**Probability/Statistics**

The answer is Prob(A) + Prob(B) - Prob (A + B)= 0.6 + 0.3 - 0.18 = 0.72 The last term avoids double counting of the occurence of both A and B, which satisfied the "A or B" criterion only once. The answer is certainly not 0.3, since 60% of the events are A and satisfy the A or ...
*Tuesday, February 24, 2009 at 2:56pm by drwls*

**Prob and Stats**

In a multiple regression with 5 predictors in a sample of 56 U.S. cities, what would be the critical value for an F test of overall significance at a= .05? A. 2.45 B. 2.37 C. 2.40 D. 2.56
*Wednesday, December 9, 2009 at 4:10pm by please help!*

**Finite! please help**

prob(tail) = .15 prob(heads) = .85 what we DON'T want is all 6 being heads prob(6 heads) = (.85)^6 prob (at least 1 tail) = 1 - .85^6 = appr .623
*Thursday, October 18, 2012 at 8:42am by Reiny*

**Prob&stats**

Search results for: Using numbers 1-20 what is probability a randomly chosen number is either an even number or a number greater than 14?
*Wednesday, February 29, 2012 at 8:58am by Rosa*

**Prob&stats**

Search results for: Using numbers 1-20 what is probability a randomly chosen number is either an even number or a number greater than 14?
*Wednesday, February 29, 2012 at 9:24am by Rosa*

**math**

prob(Ruben winning) = 16/24 = 2/3 prob(Ruben NOT winning) = 1/3 so prob(Manuel winngin) = 1/3 prob(Man not winning) = 2/3 odds in favour of Manuel winning = (1/3) : (2/3) = 1 : 2
*Thursday, April 17, 2014 at 8:46am by Reiny*

**xiamen university**

prob of cold = .62 prob of not cold = .38 a) prob of 4 of 5 catch cold = C(5,4) (.62)^4 (.38) b) prob 3 or more = prob 3 + prob 4 + prob 5 = C(5,3)(.62)^3 (.38)^2 + C(5,4) (.62)^4 (.38) + C(5,5) .62^5 = ...
*Sunday, October 28, 2012 at 2:10pm by Reiny*

**Math**

possible events: RR RB RG RY BB BG BY GG GY YY (The order does not matter) I will do one of them, you do the rest Prob(B or G) = C(3,1)*C(2,1)/C(10,2) = 6/45 = 2/15 or Prob(BG) = (3/10)(2/9) = 6/90 prob (GB) = (2/10)(3/9) = 6/90 prob (B or G) = 6/90 + 6/90 = 6/45
*Wednesday, May 2, 2012 at 8:46am by Reiny*

**math**

prob(choose math) = .65 prob(not to choose math) = .35 prob(at least 1 of 3 choosing math) = 1 - prob(nobody choosing math) = 1 - C(3,0) (.35)^3 = 1 - .042875 = appr .957 or prob(1 choosing math) + prob(2 choosing math) + prob(3 choosing math) = .957
*Tuesday, May 28, 2013 at 3:25am by Reiny*

**Statistics**

prob square = .72 then prob of triangle is 1-.72 = .28 prob(each of the particular squares) = .72/6 = .12 prob(each of the particular triangles = .28/8 = .035 note 8(.035) + 6(312) = 1
*Tuesday, December 18, 2012 at 9:18pm by Reiny*

**stats**

are you Mike? http://www.jiskha.com/display.cgi?id=1279157840 prob = C(26,5)/C(70,5) = .005435 or prob = 26/70*25/69*24/68*23/67*22/66 = .005435
*Wednesday, July 14, 2010 at 10:53pm by Reiny*

**Binomial Math**

prob of getting a 5 = 4/36 = 1/9 prob not a 5 = 8/9 prob getting a 5 twice in 4 rolls = C(4,2) (1/9)^2 (8/9)^2 = 6 (1/81)(64.81) = 128/2187 = appr .0585
*Sunday, January 27, 2013 at 2:48pm by Reiny*

**Math : Probability**

p = .5 1-p = .5 binary coefs 1 4 6 4 1 prob 0 head 4 tails = 1*.5^0*.5^4 = .0625 prob 1 head 3 tails = 4*.5^1*.5^3 = .25 prob 2 head 2 tails = 6*.5*2*.5^2 = .5625 prob 3 head 1 tail = .25 prob 4 head 0 tail = .0625 You can take it from there I think
*Monday, February 17, 2014 at 12:10pm by Damon*

**whoop, middle value miscalculated**

prob 0 head 4 tails = 1*.5^0*.5^4 = .0625 prob 1 head 3 tails = 4*.5^1*.5^3 = .25 prob 2 head 2 tails = 6*.5*2*.5^2 = .375 prob 3 head 1 tail = .25 prob 4 head 0 tail = .0625 You can take it from there I think
*Monday, February 17, 2014 at 12:10pm by Damon*

**algebra**

The prob of rolling a 6 is 1/6 half are even, half are odd faces so the prob of an even number is 1/2 so prob of the event you stated is 1/6*1/2 =1/12
*Thursday, December 6, 2007 at 1:08pm by Reiny*

**COLLEGE MATH**

There is no joint probability, that is, it cannot be consumed in China and US. Therefore, Prob(notconsumedChinaorUS)=1-prob(US)-prob(China)=.50
*Sunday, April 12, 2009 at 5:05pm by bobpursley*

**Math**

prob(Jim AND Joan) = .919(.843) = .774717 a) Prob(Jim OR Joan) = Prob(Jim) + prob(Joan) - P(Jim AND Joan) = .919 + .843 - .774717 = .987 b) prob(neither is in class) = (.081)(.157) = .0127
*Monday, January 30, 2012 at 9:48pm by Reiny*

**Probability and Statistics**

If I understand correctly, there are 5 trials and you are expecting 3 S's and 2 F's, (S = success, F = failure) defective - F -----> prob(F) = .1 non-defective - S --> prob(S) = .9 a) you want an S in the 5th spot e.g. SFFSS is one of these number of ways for your ...
*Sunday, September 29, 2013 at 3:12am by Reiny*

**Prob and Stats**

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 the Z scores calculated.
*Tuesday, October 4, 2011 at 6:16pm by PsyDAG*

**maths**

I am not familiar with your notation of ⌊1000p⌋ but I would do it this way: after a W prob(W) = 7/10 , prob(L) = 3/10 after a L, Prob(W) = 3/10, prob(L) = 7/10 To have a 4-game series in a best of 5 setup, the winning team could win in these ways LWWW --- (1/2)(3/...
*Friday, February 15, 2013 at 6:41am by Reiny*

**Math**

label your results, such as prob(red) = 20/45 = 4/9 prob(yellow) = 5/9 (don't just write down some arithmetic calculations, e.g. what is 25*20 = 500 supposed to represent ) here is all you need Prob(2 reds) = (20/45)(19/44) = 19/99 explanation: prob(first red) = 20/45 there ...
*Sunday, September 29, 2013 at 4:51am by Reiny*

**Math**

1. No of ways = 5*6*3 = 90 2. No of ways for large box = 1*6*3 = 18 so Prob(large box) = 18/90 = 1/5 (well duh, since there are 5 different sizes of boxes.....) 3. Prob(jazz bow) = 1/3 so prob(NOT a jazzy bow) = 1-1/3 = 2/3 4. prob of baby-boy = 1/6
*Tuesday, February 19, 2008 at 7:40pm by Reiny*

**stats again**

ok I get it, so once I get the prob figred for one person, how do I figure it if additional attempts are made and the question is about the prob of at least one of the additional attempts say n = 6 getting the same results as the first Once you have the probability of passing ...
*Thursday, July 5, 2007 at 7:38pm by holly*

**Probability**

Agree that the numbers are enourmous, so I will leave answers in notation form a) prob of having ticket = 19/20 So prob that 200 people have a ticket = (19/20)^200 b) prob that at most 3 will have no tickets = prob(3 no ticket) + prob(2 no ticket) + prob(1 no ticket) + prob(0 ...
*Friday, April 30, 2010 at 2:09am by Reiny*

**Prob Stats**

Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
*Monday, January 9, 2012 at 7:55am by PsyDAG*

**Prob&stats**

Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
*Wednesday, February 29, 2012 at 9:24am by PsyDAG*

**MATH 12 HELP!**

Since a coin toss is used to select either bag1 or bag2 the prob that bag1 is choses is 1/2 prob that a white ball is choses from bag1 = 4/10 = 2/5 so prob of your 'event' = (1/2)(2/5) = 1/5
*Sunday, May 20, 2012 at 2:15am by Reiny*

**Math**

odds that the horse will lose = 5 : 3 so the prob it will lose is 5/8 and the prob it will win is 3/8 odds in favour of some event happening = prob(the event will happen) : prob(the event will not happen)
*Wednesday, January 27, 2010 at 7:03pm by Reiny*

**math**

there are only 4 cases: WW --- prob is (3/5)(2/4) = 6/20 WR --- prob is (3/5)(2/4) = 6/20 RW --- prob is (2/5)(3/4) = 6/20 RR --- prob is (2/5)(1/4) = 2/20 so different colours are : WR and RW = 6/20 + 6/20 = 12/20 = 3/5 (notice the 4 cases add up to 1)
*Monday, December 27, 2010 at 12:18am by Reiny*

**Word Problem.**

Prob (HHH) = 1/8 expected value = (1/8)(15) = 1.875 prob(HHT or HTH or THH) 3/8 expected winnings = (3/8)(5) = 1 .875 prob(HTT or THT or TTH) = 3/8 expected winnings = (3/8)(2) = .75 prob( TTT) = 1/8 expected winning = 0 total expected winnings = 4.5 to play the game costs $4 ...
*Sunday, March 7, 2010 at 8:40pm by Reiny*

**math**

the probabilities that 3 friends A,B andC pass a driving test are1/3,1/4,2/5 respectively.All 3 take the test find the probability that(a)aa 3 failed the test (b) only B passes the test(c) only 2 of them pass the test (d)at least 1 passes the test... I will do one. Assuming ...
*Tuesday, June 26, 2007 at 9:17am by celina*

**math**

the prob you will pick a correct first number = 4/10 the prob that the second one is correct = 3/9 etc so the prob of drawing a 1,2,3, and 4 = (4/10)(3/)(2/8)(1/7) = 1/210 or total number of ways of 'choosing' 4 specifics form 10 is C(10,4) = 210 only one of these will be the ...
*Thursday, April 9, 2009 at 11:49am by Reiny*

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