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The probability that a visitor at the mall buys a pack of [#permalink]
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16 Nov 2007, 09:27
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The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy? A. .343 B. .147 C. .189 D. .063 E. .027
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(c)
Probablity that two buy candy is:
(0.3)(0.3)(0.7) = 0.063
Three ways that this can happen
A+B
A+C
B+C
3 * 0.063 = 0.189



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If the question asked for the probability that customer 1 and customer 3 bought candy, the answer would be:
(.3)(.7)(.3) = .063
Is this reasoning right? Similar to the question about rain on the first 2 days of the week given a probability for rain.



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CaspAreaGuy wrote: Guys, will the answer be different if the question asked for a probablity that at least two will buy candies? Can anyone please explain?
Yes, right now the question says exactly two. So we need to find the probability that EXACTLY 2 people buy candy. If it said at least two, then we need to find the probability that 2 OR 3 people bought candy.
This would result in the answer we have for 2 people buying candy PLUS the probability of all 3 people buying candy.
(.3)(.3)(.7)*3 = .189
(.3)(.3)(.3)*1 = .027
.189 + .027 = .216 probability of at least 2 people buying candy.



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Re: PS Candy Probability [#permalink]
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22 Aug 2008, 08:46
yuefei wrote: The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?
a. .343 b. .147 c. .189 d. .063 e. .027 0.3*0.3*0.7 + 0.3*0.3*0.7 + 0.3*0.3*0.7 = 0.189
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Re: PS Candy Probability [#permalink]
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04 Oct 2008, 11:22
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•P(Buy) = 3/10, P(No Buy) = 7/10 •2 Yes, 1 No = 3/10*3/10*7/10 = 63/1000 •3 Possibilities = YYN + YNY + NYY = 3(63/1000) = 189/1000 = .189%



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eschn3am wrote: CaspAreaGuy wrote: Guys, will the answer be different if the question asked for a probablity that at least two will buy candies? Can anyone please explain? Yes, right now the question says exactly two. So we need to find the probability that EXACTLY 2 people buy candy. If it said at least two, then we need to find the probability that 2 OR 3 people bought candy. This would result in the answer we have for 2 people buying candy PLUS the probability of all 3 people buying candy. (.3)(.3)(.7 )*3 = .189 (.3)(.3)(.3) *1 = .027 .189 + .027 = .216 probability of at least 2 people buying candy. can someone explain why should we multiply by three and one, thanks in advance



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kazakhb wrote: eschn3am wrote: CaspAreaGuy wrote: Guys, will the answer be different if the question asked for a probablity that at least two will buy candies? Can anyone please explain? Yes, right now the question says exactly two. So we need to find the probability that EXACTLY 2 people buy candy. If it said at least two, then we need to find the probability that 2 OR 3 people bought candy. This would result in the answer we have for 2 people buying candy PLUS the probability of all 3 people buying candy. (.3)(.3)(.7 )*3 = .189 (.3)(.3)(.3) *1 = .027 .189 + .027 = .216 probability of at least 2 people buying candy. can someone explain why should we multiply by three and one, thanks in advance Three conditions in which two people can buy....12, 23 or 13. Only one condition in which all three people can buy...123. Hence, the first probability is multiplied by 3 whereas the second probability is multiplied by only 1.



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Re: PS Candy Probability [#permalink]
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24 Jul 2009, 19:13
Since the three people are DISTINCT thts why the anser is .063x3=.189:)
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Re: PS Candy Probability [#permalink]
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12 Sep 2009, 18:44
Could someone please let me know where I've messed up my calculation?
P(exactly 2 visitors buying candy) = 1  P(exactly 3 visitors buying candy)  P(exactly 1 visitor buying candy)  P(no visitors buying candy) = 1  (3/10)^3  3/10*7/10*7/10  (7/10)^3 = 1  0.027  0.147  0.343 = 0.483
Many thanks!



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Re: PS Candy Probability [#permalink]
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pinktyke wrote: Could someone please let me know where I've messed up my calculation?
P(exactly 2 visitors buying candy) = 1  P(exactly 3 visitors buying candy)  P(exactly 1 visitor buying candy)  P(no visitors buying candy) = 1  (3/10)^3  3/10*7/10*7/10  (7/10)^3 = 1  0.027  0.147  0.343 = 0.483
Many thanks! P(exactly 1 visitor buying candy) = 3 * 3/10*7/10*7/10 = .441 (Between A, B & C it could be A or B or C) or, P(exactly 2 visitors buying candy) = 1  0.027  0.441  0.343 = .189



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Re: PS Candy Probability [#permalink]
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12 Sep 2009, 19:17
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pinktyke wrote: Could someone please let me know where I've messed up my calculation?
P(exactly 2 visitors buying candy) = 1  P(exactly 3 visitors buying candy)  P(exactly 1 visitor buying candy)  P(no visitors buying candy) = 1  (3/10)^3  3/10*7/10*7/10  (7/10)^3 = 1  0.027  0.147  0.343 = 0.483
Many thanks! The possibility that exactly 1 visitor buying candy is three times you calculated. This is because positioning. 100, 010, 001 10,0270,147*30,343 =10,0270,4410,343 =10,811 =0,189



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Re: PS Candy Probability [#permalink]
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27 Sep 2009, 02:27
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?
a. .343 b. .147 c. .189 d. .063 e. .027
Soln:
Probability that exactly two will buy is = P(First Buys, Second Buys,Third does not buy) + P(First buys, Second does not buy,Third buys) + P(First does not buy, Second Buys,Third Buys) = (3/10 * 3/10 * 7/10) * 3 = .189



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Re: PS Candy Probability [#permalink]
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09 Sep 2010, 21:31
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yuefei wrote: The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?
a. .343 b. .147 c. .189 d. .063 e. .027 Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189 Answer: C. Explanation: 3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB > =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t. NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t. Let’s consider some similar examples:1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy? The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did. So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441 NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did. 2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?At least ONE buys, means that buys exactly one OR exactly two OR exactly three: P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657 P(B=1) > 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s) P(B=2) > 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s) P(B=3) > 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy  BBB. BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below: P(B>=1)=1P(B=0, the same as N=3)=13!/3!*0.7^3=10.7^3. 3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?P(B=2)=5!/2!3!*0.3^2*0.7^3 We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) > 2 identical Bs and 3 identical Ns, total of five visitors > 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 Bs=0.3^2 and 3 Ns=0.7^3. Also discussed at: probability85523.html?hilit=certain%20junior%20class#p641153Hope it helps.
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Re: The probability that a visitor at the mall buys a pack of [#permalink]
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23 Feb 2012, 15:07
The binomial probability formula seems like overkill for this, but I like to use it when I can so I can remember how to use it... This page explains the Binomial Probability formula: http://stattrek.com/lesson2/binomial.aspxQuote: Suppose a binomial experiment consists of \(n\) trials and results in \(x\) successes. If the probability of success on an individual trial is \(P\), then the binomial probability is: \(b(x; n, P) = nCx * P^x * (1  P)^{n  x}\)
In this problem: n=3 x=2 p=3/10 \(b(x;n,p) = 3C2 * (3/10)^2 * (7/10)\) = \({3 * 3 * 3 * 7} / 1000\) = \(.189\) An alternate approach: S implies Success, F implies Failure \(P(exactly two successes) = P (SSF) + P (SFS) + P (FSS)\) \(= (3/10 * 3/10 * 7/10) + (3/10 * 7/10 * 3/10) + (7/10 * 3/10 * 3/10)\) \(= 3 * (3 * 3 * 7 / 1000)\) \(= .189\)



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Re: PS Candy Probability [#permalink]
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14 May 2014, 21:40
Bunuel wrote: yuefei wrote: The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?
a. .343 b. .147 c. .189 d. .063 e. .027 Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189 Answer: C. Explanation: 3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB > =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t. NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t. Let’s consider some similar examples:1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy? The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did. So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441 NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did. 2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy? At least ONE buys, means that buys exactly one OR exactly two OR exactly three: P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657 P(B=1) > 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s) P(B=2) > 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s) P(B=3) > 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy  BBB. BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below: P(B>=1)=1P(B=0, the same as N=3)=13!/3!*0.7^3=10.7^3. 3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy? P(B=2)=5!/2!3!*0.3^2*0.7^3 We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) > 2 identical Bs and 3 identical Ns, total of five visitors > 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 Bs=0.3^2 and 3 Ns=0.7^3. Also discussed at: probability85523.html?hilit=certain%20junior%20class#p641153Hope it helps. You said that probabilty of atleast 1 = 1  probabiliy of 0, but won't probability of atleast 1 = probability of atmost 1? im a little confused as to how probablity of atleast 1 = probability of 0. Please help me with this



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Re: PS Candy Probability [#permalink]
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15 May 2014, 01:37
havoc7860 wrote: Bunuel wrote: yuefei wrote: The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?
a. .343 b. .147 c. .189 d. .063 e. .027 Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189 Answer: C. Explanation: 3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB > =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t. NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t. Let’s consider some similar examples:1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy? The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did. So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441 NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did. 2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy? At least ONE buys, means that buys exactly one OR exactly two OR exactly three: P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657 P(B=1) > 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s) P(B=2) > 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s) P(B=3) > 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy  BBB. BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below: P(B>=1)=1P(B=0, the same as N=3)=13!/3!*0.7^3=10.7^3. 3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy? P(B=2)=5!/2!3!*0.3^2*0.7^3 We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) > 2 identical Bs and 3 identical Ns, total of five visitors > 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 Bs=0.3^2 and 3 Ns=0.7^3. Also discussed at: probability85523.html?hilit=certain%20junior%20class#p641153Hope it helps. You said that probabilty of atleast 1 = 1  probabiliy of 0, but won't probability of atleast 1 = probability of atmost 1? im a little confused as to how probablity of atleast 1 = probability of 0. Please help me with this I guess you are talking about example #2. 2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?At least 1 visitor buys out of 3, means 1, 2, or all 3 visitors buy, so all the cases but when noone buys (while at most 1 out of 3 means 0 or 1). Hence the probability that at least 1 visitor buys out of 3 = 1  (the probability that noone buys). Does this make sense?
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Re: PS Candy Probability [#permalink]
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15 May 2014, 01:40
havoc7860 wrote: Bunuel wrote: yuefei wrote: The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?
a. .343 b. .147 c. .189 d. .063 e. .027 Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189 Answer: C. Explanation: 3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB > =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t. NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t. Let’s consider some similar examples:1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy? The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did. So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441 NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did. 2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy? At least ONE buys, means that buys exactly one OR exactly two OR exactly three: P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657 P(B=1) > 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s) P(B=2) > 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s) P(B=3) > 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy  BBB. BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below: P(B>=1)=1P(B=0, the same as N=3)=13!/3!*0.7^3=10.7^3. 3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy? P(B=2)=5!/2!3!*0.3^2*0.7^3 We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) > 2 identical Bs and 3 identical Ns, total of five visitors > 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 Bs=0.3^2 and 3 Ns=0.7^3. Also discussed at: probability85523.html?hilit=certain%20junior%20class#p641153Hope it helps. You said that probabilty of atleast 1 = 1  probabiliy of 0, but won't probability of atleast 1 = probability of atmost 1? im a little confused as to how probablity of atleast 1 = probability of 0. Please help me with this Some "at least" probability questions to practice: leilaisplayingacarnivalgameinwhichsheisgiven140018.htmlafaircoinistossed4timeswhatistheprobabilityof131592.htmlforeachplayersturninacertainboardgameacardis132074.htmlastringof10lightbulbsiswiredinsuchawaythatif131205.htmlashipmentof8tvsetscontains2blackandwhitesetsand53338.htmlonashelfthereare6hardbackbooksand2paperbackbook135122.htmlinagroupwith800people136839.htmltheprobabilityofamanhittingabullseyeinonefireis136935.htmlforeachplayersturninacertainboardgameacardis141790.htmltheprobabilitythataconveniencestorehascansoficed128689.htmltripletsadambruceandcharlieenteratriathlonif132688.htmlamanufacturerisusingglassasthesurface144642.htmltheprobabilityis12thatacertaincoinwillturnuphead144730.html (OG13) afaircoinistobetossedtwiceandanintegeristobe148779.htmlinagameoneplayerthrowstwofairsixsideddieatthe151956.htmlHope it helps.
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