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Old 07-27-2007   #1
nguardwife
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Default Probability Question?

An experiment consists of rolling a pair of dice 10 times. On each roll the sum of the dots on the two dice is noted.Find the probability that in the 10 rolls of the pair of dice, a 7 or 11 occurs 5 times.
 
Old 07-27-2007   #2
Brad K
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There are 36 combinations, of which, 6 are equal to seven and 2 are equal to eleven.Odds of getting seven or eleven are: 2/9Odds of NOT getting a seven or eleven are: 7/9To get exactly 5 sevens, 2/9*2/9*2/9*2/9*2/9(for the five sevens) and 7/9*7/9*7/9*7/9*7/9 (for the 5 NON-sevens).(2^5 * 7^5) / (9^10)You should be able to get the answer from there.
 
Old 07-27-2007   #3
raja
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Sample space is 6^10,Odds of getting seven are: 1/6Odds of NOT getting a seven are: 5/6The number of ways you can hit a 7 or 11 is (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), and (6, 5). So the probability of hitting would be 8/36 or 2/9.So the probability of hitting this exactly 5 times in 10 rolls would be C(10, 5) * (p)^5 (1-p)^5, where C(10, 5) is "10 choose 5", and p = 2/9. Evaluating this we get 252*[(2/9)^5](7/9)^5] = 0.03887.
 
Old 07-27-2007   #4
Patrick W
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your odds of rolling a seven on two dice = 6/36your odds of rolling an eleven on two dice = 2/36 P(7 occurs five times)= (6/36)^5 * (30/36)^5P (11 occurs five times) = (2/36)^5 * (34/36)^5
 
Old 07-27-2007   #5
Armadura
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36 combinations,6 are equal to seven and 2 are equal to eleven.Odds of getting seven or eleven are: 8/36=2/9Odds of not getting a seven or eleven are: 7/9The number of seven or eleven is a Binomial with n=10 and p=2/9 thenP(X=5)=10C5(2/9)^5(7/9)^510C5=(10*9*8*7*6*5!)/(5!*(10-5)!)=10*9*8*7*6/(5*4*3*2)=2*3*2*7*3=252P(X=5)=252*(32/59049)*(3125/59049)=0.007227289
 
Old 07-27-2007   #6
Steiner
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Lets's define the eventsA= {7 ocurrs 5 times} and B = {11 occurs 5 times}So, what we want is P(A U B) (probabilty of the union of A and B). As we know, P(A U B) = P(A) +P(B) - P(A intersect B).On each roll, either the sum is 7 or it's not 7. We have 36 possibilities and, in order to get sum 7, we must get one of the following pairs: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). So, on each roll, there are 6 in 36 cases favorable to the event {the sum is 7}, and this implies the probability of sum =7 is 6/36 = 1/6. Since the 10 rolls are independent of each other, we have a binomial distribution, with parameters n =10 (number of rolls) and p = 1/6 (probailty of getting sum 7. According to the formula for binomial distribution, it the follows that P(A) = C(10, 5)(1/6)^5(1/6)5, where C(10,5) is 10 choose 5. So, P(A) =~ 0.01302381000The evaluation of P(B) is completely similar. Now we have 2 cases favorable to sum =11, namely (5,6) and (6,5) so that the probability of getting sum 11 in each roll is 2/36 = 1/18. It follows P(B) = C(10,5)(1/18)^5(1/18)^5 =~ 0.000100212Now, it remains to compute P(A intersect B), which means that both sum = 7 and sum =11 occur 5 times. Since the rolls are independent, the probability that a a particular sequence satisfies such condition is (1/6)^5 * (1/18)^5. To find the number of such sequences, we choose the order of those where 7 will occur, and the order of those where 11 will ocur gets automatically defined. So, we have C(10, 5) sequences satisfying the desired condition, and it folows P(A U B) = C(10,5)(1/6)^5(1/18)^5 =~ 1.71507E-08, a very small number.Finally, our answer is P(A U B) = P(A) + P(B) - P(a intersect B) = 0.01302381000 + 0.000100212 - 1.71507E-08 = ~ 0.013124005 or 1.312400514%
 
 

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