Friday

April 18, 2014

April 18, 2014

Number of results: 214,368

**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*

**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*

**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*

**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*

**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*

**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**

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**

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*

**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*

**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*

**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*

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

**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*

**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*

**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*

**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*

**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*

**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*

**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**

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-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*

**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*

**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*

**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 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*

**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*

**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*

**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*

**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*

**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 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*

**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*

**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*

**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*

**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*

**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**

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*

**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*

**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*

**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*

**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*

**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*

**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*

**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 : 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*

**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*

**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 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**

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*

**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*

**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*

**math**

odds in favour of some event = prob(event)/prob(not event) so prob of pat winning = 7/13 check: prob pat will win = 7/13 prob pat will lose = 6/13 odds in favour of losing = (6/13) / (7/13) = 6/7 or 6 : 7
*Saturday, March 10, 2012 at 9:51pm 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*

**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*

**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*

**math, pre-statistics**

Prob(rain 1stday)*prob(rain 2nd day)*prob(rain 3rd day) = (.6)(.6)(.6) = .216
*Sunday, February 8, 2009 at 9:22pm by Reiny*

**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**

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**

Ah, a variation on the good old St Petersburg paradox. Prob of winning $1 is .5 Prob of winning $2 is .5*.5 Prob of winning $4 is .5*.5*.5 Prob of winning $8 is .5*.5*.5*.5 and so on, Expected value is sum over all possible outcomes, the probability times the value of the ...
*Sunday, November 2, 2008 at 3:32pm by economyst*

**finite math**

d - defective g -good a) prob(ddd) = (5/40)(4/39)(3/38) = .... b) prob(ggg) = (35/40)(34/39)(33/38) = .... c) prob(dgg) + prob(gdg) + prob(ggd) = 3(5/40)(35/39)(34/38) = ... d) 1 - prob(ddd) = 1 - (35/40)(34/39)(33/38)
*Thursday, March 24, 2011 at 1:09pm by Reiny*

**math**

why not just find the prob of getting a sum of 7 ? there are 6 ways to get a 7 (list them if you have to) prob of getting a sum of 7 = 6/36 = 1/6 btw. prob of 11 : 2 ways, either 5,6 or 6,5 prob of 11 = 2/36 = 1/18 , as they stated and (3)(1/18) = 1/6 (they just made the ...
*Wednesday, May 2, 2012 at 11:42am by Reiny*

**Math**

prob of picking the right key = 2/5 prob of opening on first try = 2/5 prob of opening on 2nd try = (3/5)(2/5) = 6/25 (2nd try has to be Miss, Success) prob of 1st OR 2nd try = 2/5 + 6/25 = 16/25
*Saturday, November 24, 2012 at 1:51pm by Reiny*

**math**

odds and probability are not quite the same the odds in favour of some event = (the prob of that even will happen):(prob that the event will NOT happen) so prob of 7 or 12 = 7/36 so prob of not a 7 or 12 is 29/36 so the odd of getting a 7 or 12 is 7:29 (if the odds of some ...
*Wednesday, September 23, 2009 at 12:50am 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*

**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**

a) The prob that a guess is wrong is 3/4 so the prob that none are correct, that is, all 3 are wrong is (3/4)(3/4)(3/4) = 27/64. b) then the prob that all 3 are guessed correctly is (1/4)(1/4)(1/4) = 1/64
*Sunday, August 16, 2009 at 1:43pm by Reiny*

**Math (probability) repost for Christine**

The prob. that she will lose is 5/7 the prob. that she will win is 2/7
*Tuesday, October 16, 2007 at 9:48am by Reiny*

**Math**

Prob(B given A)=Prob(A,B)/Pr(A) = .02/.08=.25
*Friday, November 6, 2009 at 9:00pm by bobpursley*

**math/probability**

a) prob (red,red) = (1/4)(1/4) = 1/16 b) prob(red then green) = (1/4)(1/4) = 1/16 c) could be RG or GR so prob = 1/16 + 1/16 = 1/8 (I don't see where the "at least" comes in, you are only drawing 2 and you want a red AND a green) d) prob (no yellow, no yellow = (3/4)(3/4) = 9/16
*Wednesday, October 21, 2009 at 10:43pm by Reiny*

**EASY MATH QUESTION**

So the only case you don't want is a 3 which has a prob of 1/6 so prob(of your stated event) = 5/6
*Tuesday, June 7, 2011 at 2:38pm by Reiny*

**math-Probablity**

Since you want the prob of making 23 or better, you would want the prob that she makes 23 of 27 +24 of 27 + 25 of 27 + 26 of 27 + 27 of 27 you will have 5 prob's to calculate , add them up
*Tuesday, June 5, 2012 at 4:10pm by Reiny*

**math**

a) prob = (2/5)(1/4) = 2/20 = 1/10 b) prob = (2/5)(2/5) = 4/25
*Wednesday, November 18, 2009 at 9:34pm by Reiny*

**math**

prob(defective) = 4/12 = 1/3 prob(not defective) = 2/3 1. all 3 defective ---- prob = (1/3)^3 = 1/27 2. at least 2 defective ----> 2 defective or 3 defective prob = C(3,2)(1/3)^2 (2/3) + (1/3)^3 = 3(2/27) + 1/27 = 7/27 3. at most 2 ---> 0 defective, 1 defect, or 2 ...
*Sunday, September 23, 2012 at 10:38pm by Reiny*

**math**

odds in favour of some event = prob(of that event/prob(not that event) primes form 1 to 6 are 2,3, and 5 so prob of a prime = 3/6 = 1/2 so prob of not a prime = 1-1/2 = 1/2 so the odds in favour of a non-prime = (1/2) ÷ (1/2) = 1:1
*Tuesday, June 16, 2009 at 11:03am by Reiny*

**MATH Prob.**

same argument as the previous question... find prob of at least one head = 7/8 etc
*Tuesday, August 11, 2009 at 11:42pm by Reiny*

**math**

Prob(E) = 1/8 prob(notE) = 7/8 odds against the event = (7/8) : (1/8) = 7 : 8 do the other one the same way.
*Monday, May 30, 2011 at 7:31pm by Reiny*

**Math**

first one: C(6,1)*C(2,1)/C8,2) = 6*2/28 = 3/7 second one: C(5,2)*C(4,2)/C(16,4) = 10*6/1820 = 3/91 third one: the key word is the "or" prob(not red OR not a facecard) = Prob(not red) + prob(not a facecard) - prob(not red AND not a facecard) = 26/52 + 40/52 - 20/52 = 46/52 = 23...
*Thursday, April 24, 2008 at 11:09pm by Reiny*

**Math**

prob (hit) = 42/60 = 7/10 prob(miss) = 3/10 prob 0 points = 3/10 , (only gets one shot, a miss) prob 1 point = (7/10)(3/10) = 21/100 , (has to hit first, miss the second prob 2 points = (7/1)(7/10) = 49/100 (has to make them both did you notice that 3/10 + 21/100 + 49/100 = ...
*Wednesday, December 14, 2011 at 9:52pm by Reiny*

**Math probability**

4+7+3+2 = 16 socks prob red first = 4/16 prob green second = 7/15 prob that happens = (4*7)/(16*15) prob green first = 7/16 prob red second = 4/15 so again (4*7)/(16*15) so in the end 2 * 4 * 7 / (15*16) = .233
*Monday, March 31, 2014 at 5:38pm by Damon*

**math**

take prob that they are both different first can be anything, the second must be one of the other 54 prob that they are both different = (1)(54/55) = 54/55 prob( they are NOT both different) = prob(at least two are the same) = 1 - 54/55 = 1/55
*Monday, October 22, 2012 at 1:53am by Reiny*

**math**

prob of same (HH, TT) = (.6)(.6) + (.4)(.4) = .36+.16 = .52 prob of different (HT, TH) = 2(.6)(.4) = .48
*Wednesday, July 14, 2010 at 5:41pm by Reiny*

**Math**

from your experiment, the empirical prob of a 5 = 8/48 = 1/6 (btw, this matches the actual prob of rolling a 5 for a true die)
*Monday, May 27, 2013 at 10:11am by Reiny*

**Math**

from your experiment, the empirical prob of a 5 = 8/48 = 1/6 (btw, this matches the actual prob of rolling a 5 for a true die)
*Monday, May 27, 2013 at 10:11am by Reiny*

**Math**

How do I work this math prob. I always have a B and would like to bring it up. Caan you PLEASE help me with this prob. on unit rates "$2 for 5 cans of tomato soup"
*Monday, January 10, 2011 at 9:11pm by Sarah*

**Math ASAP please**

in #1 - 3 you don't state what the event is, so those questions are incomplete #4. Odds in favour of some event = prob of the event / prob of not the event prob of correct choice = 1/5 so the odds in favour of picking the right answer = (1/5 / 4/5) = 1/4 or 1 : 4
*Friday, March 29, 2013 at 9:59pm by Reiny*

**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*

**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*

**math**

Your question is incomplete. Are you choosing 2 cars ? It doesn't say. Assume we are choosing 2 cars and you want the prob that they are both of the same colour. The could be RR or WW or BB prob(RR) = C(50,2)/C(100,2) = 1225/4950 = 49/198 prob(WW) = C(30,2)/C(100,2) = 435/4950...
*Wednesday, January 16, 2013 at 12:33pm by Reiny*

**Math**

prob of getting one black from 1st box = 3/4 prob of getting one black from 2nd box = 6/8 = 3/4 mmmmmhhhh? prob(2 blacks from 1st) = (3/4)(2/3) = 1/2 prob(2 blacks from 2nd) = (6/8)(5/7) = 15/28 mmmh again!
*Tuesday, November 12, 2013 at 10:25am by Reiny*

**MATH**

Prob(blue eyes) = .2 Prob(not blue eyes)= .8 prob( at least 4 of 5 have blue eyes) = prob(exactly 4 of 5 have blue eyes) + prob(exactly 5 of 5 have blue eyes) = C(5,4)(.2)^4 (.8) + C(5,5)(.2)^5 = 5(.0016)(.8) + 1(.00032) = .000672
*Friday, March 25, 2011 at 12:44am by Reiny*

**math**

So what we DON"T want is the case of "all even" Let's find the prob of that. prob of 3 all even = (4/9)(3/8)(2/7) = 1/27 So prob of at least one odd = 1 - 1/27 = 26/27
*Monday, September 13, 2010 at 10:22pm 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*

**Math-Probability**

1. prob = (1/6)(1/6) = 1/36 2. the colour does not matter there are 4 ways to get a sum of 5 (1,4) (2,3) (3,2) (4,1) prob = 4/36 = 1/9 3. Again, what does the colour have to do with it? prog(no 6 showing) = (5/6)(5/6) = 25/36 prob(at least one 6) = 1 - 25/36 = 11/36
*Thursday, November 14, 2013 at 7:01pm by Reiny*

**math**

If machine is working on day 0 prob working on day 1 = .9 prob broken on day 1 = .1 If machine is broken on day 0 prob working on day 1 = .8 prob broken on day 1 = .2 so working broken 1 = working broken 0 * .9 .1 .8 .2 for example if working day 0 (doing second row with / ...
*Sunday, February 15, 2009 at 10:10am by Damon*

**math/probability**

let A be "sum of 6" and B be "double" What you have is P(A│B) , read as the probability of A given B which is defined as P(A AND B)/P(B) Prob(A AND B) = Prob(sum of 6 AND a double) = 1/36 Prob(B) = 6/36 = 1/6 so Prob(sum of 6 AND a double) = (1/36)÷(1/6) = (1/36)(6/1) = 1/6
*Monday, December 7, 2009 at 12:32am by Reiny*

**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*

**math**

toss the coin once, what is the prob of getting heads? 1/2 right? now does the coin have a memory of what happened last time? so toss it again. What is the prob this time? Still 1/2, right? now now you have 1/2*1/2 of two in a row. What would it be 3 times in a row??? BTW, a ...
*Monday, December 15, 2008 at 5:31pm by Reiny*

Pages: **1** | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>