Saturday
April 19, 2014

Search: Prob/Stat

Number of results: 1,721

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

prob and stat (?)
Sunday, October 30, 2011 at 9:24am by PsyDAG

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

Prob/Stat
If P(A) = .3 and P(B) = .4 and P(AandB) = .2 what is P(A/B)? P(B/A)? Are A and B indepentent?
Sunday, April 1, 2012 at 8:44am by Sheila

math/prob stat
Gucci
Tuesday, March 30, 2010 at 8:42pm by Anonymous

Prob ans stat
Friday, July 29, 2011 at 12:19pm by Tanesha

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

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

Math...For All Math Tutors
One of my three sons was a math/econ undergrad major from U Cal Santa Barbara. He found it difficult to get a job at first, then took additional actuary stat/prob correspondence courses and got certified as an actuary. That landed him a great job at Social Security ...
Friday, February 15, 2008 at 10:29am by drwls

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

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

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

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

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

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

graphing
I put the numbers in using the stat function of the TI-83. I hit STAT and edit to put in the numbers in L1 and L2. Then I did Stat -Calc and I chose Quadreg and L1 L2 from keyboard and enter.
Friday, February 22, 2013 at 12:29pm by JJ

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

probability and statistics.
So here I am trying to do my prob/stat homework and I keep coming across this word that I don't understand. The word is "inferior". My book has no glossary and I cannot find a definition that makes sense.Can somebody help me understand?
Thursday, January 27, 2011 at 4:45pm by penny lane

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

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

statistics
survey revealed two percent of students did not select stat. for biz. suppose we select a sample of 10 students. -what is the probability that 3 students selected stat. for biz? -how many students would you expect to select stat.for biz?
Monday, March 8, 2010 at 10:47am by nora

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

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

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(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

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

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

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

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

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

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
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

Probability/Statistics
1 and 2 are correct for 3 I got 23/196 there are 25 tickets greater than 25 so prob = 25/50 x 24/49 x 23/48 = 23/196 for 4 you have to use a formula that says: P(A or B) =P(A) + P(B) - P(A and B) since the prob of them being 3 yellow AND 3 blue is zero P(3yellow) = 4/12 x 3/11...
Thursday, February 7, 2008 at 10:17am by Reiny

Statistics and Probability
prob of causing illness = .3 prob of not causing illness = .7 so you want prob(7of 12will cause illness) + prob(8of 12will cause illness)+ ... + prob(12of 12will cause illness) = C(12,7)(.3)^7(.7)^5 + C(12,8)(.3^8)(.7^4) + ... +C(12,12)(.3^12) you do the button pushing.
Sunday, January 2, 2011 at 10:23pm by Reiny

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

pre-calculus
A) prob (4reds) = (7/12)(6/11)(5/10)(4/9) = 7/99 or Prob = C(7,4)/C(12,4) = 35/495 = 7/99 B) could be GGGG, GGGR Prob = C(5,4)/C(12,4) + C(5,3)*C(7,1)/C(12,4 = 5/495 + 70/495 = 75/495 = 5/33 C) can't have all red or all green prob = 1 - (all red + all green) number of all ...
Monday, May 16, 2011 at 7:45pm 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
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

Statistics
c) could be RBY RYB YBR YRB BRY OR BYR prob of each one is (1/6)(3/5)(2/4) = 6/120 = 1/20 the others come out to the same result so prob (all different) = 6/20 = 3/10 d) prob = C(4,2) (1/6)^2 (5/6)^2 = 6(1/36)(25/36) = 25/216
Wednesday, January 12, 2011 at 10:27pm by Reiny

Probability
I will have to disagree with Jen. The odds in favour of some event is defined as prob(that event) : prob(not that event) since prob(red) = 4/11 then prob(not red) = 7/11 so odds against red = (7/11) : (4/11) = 7 : 4
Tuesday, October 12, 2010 at 11:58am 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
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
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, 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 2
prob(2 on 1st, 4 on 2nd) = (1/4)(1/4) = 1/16 prob(both a 3) = (1/4)(1/4) = 1/16 third question: 3 on one, one less on the other ??? do you mean (3,2) and (2,3) if so , then prob = 1/16 + 1/16 = 1/8
Wednesday, February 1, 2012 at 11:55am 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/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

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

Probability Pțease helpp
If a and b are decimals between 0 and 1, then A and B could be either 0 or 1 The prob of a rounded up to 1 is 1/2 and the prob of a becoming 0 is 1/2 forming a table: A B -----C 0 0 ----- 0 e.g. .2+.1 = .3 --> 0 0 0 ----- 1 e.g. .4 + .4 = .8 ---> 1 ...... FALSE 0 1...
Thursday, November 24, 2011 at 6:55am by Reiny

probability question
odds and probability are not the same thing, even though they are closely related. if the odds in favour of some event are a : b then the prob of that event = a/(a+b) so if the odds against you winning are 1 : 7 then the prob of losing = 1/8 and the prob that you will win = 1...
Sunday, April 8, 2012 at 5:16pm by Reiny

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