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 Probability Old Exam P Forum

#1
05-04-2019, 10:44 AM
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20
One problem in SOA study manual

Hi everybody, I am confused by this problem:

I think, in terms of calculating the probability, we should not use combination, rather, we should use permutation, because the order actually matters (picking QND counts differently from picking DNQ), so my solution would be

total: 6*5*4 = 120

one Q: 4*3 = 12

two Q: 4

probability would be (12*2+4)/120

what is wrong with my thoughts? thanks~
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#2
05-04-2019, 11:09 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,285

You need to explain yourself. Why does order matter, and where did your calculations except 6*5*4 come from?

You could use permutations, with 120 as the denominator, as long as you use the correct corresponding numerator. There is no need or advantage to using permutations.
#3
05-04-2019, 11:32 AM
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20

Quote:
 Originally Posted by Gandalf You need to explain yourself. Why does order matter, and where did your calculations except 6*5*4 come from? You could use permutations, with 120 as the denominator, as long as you use the correct corresponding numerator. There is no need or advantage to using permutations.
I don't know how to express my thoughts, but I feel like combination is not correct, order really matters in this case.
#4
05-04-2019, 11:42 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,285

If you want to use order, write down all the possible orders that work, and then calculate how many ways there are of getting that order. You would then conclude that the combination method works.

For example, Qnd works, and there are 2*3*1 ways of getting that order.
#5
05-04-2019, 11:56 AM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,924

For this particular problem the easiest way is to calculate the probability of the complement which is the probability of not picking a quarter. You could do this by multiplying probabilities (4/6)(3/5)(2/4) or by using a counting rule.
#6
05-06-2019, 03:47 PM
 actuaryxjh SOA Join Date: Aug 2018 College: Rutgers University Posts: 20

Thank you Gandalf and Academic Actuary~