 XP Math - Forums - View Single Post - Probability Question?
 Thread: Probability Question? View Single Post 07-27-2007 #6 Steiner Guest   Posts: n/a 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%