Have a roll (2)!
Three unbiased 11-sided dice are thrown.
What are the chances, expressed as B A , that the rolls return results that fulfills BOTH the following criteria:
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The rolls have an odd numbered product,
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There is at least one repeated number.
Find B-A.
Try out my other Probability Problems !
Have a roll (1) ! is a stepping stone to this problem.
The answer is 1235.
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3 solutions
Lets denote the possibilities by a , b , c .
Total number of cases = 1 1 3
To fulfill ( 1 ) all of a , b , c must be odd because if one of them becomes even, the product will be even which is not desired.
To fulfill ( 2 ) we have two types ( a , a , b ) and ( a , a , a )
Type1: ( a , a , b )
We need to choose one odd number from 1 , 3 , 5 , 7 , 9 , 1 1 and call it ′ a ′ and one odd number from remaining ′ 5 ′ odd numbers and call it ′ b ′ .
This can be done in ( 1 6 ) ⋅ ( 1 5 ) ways
( a , a , b ) can be permuted among themselves in 2 ! 3 ! = 3 ways.
Total number of ( a , a , b ) type ordered pairs = 2 ! 3 ! ( 1 6 ) ⋅ ( 1 5 ) = 3 × 6 × 5 = 9 0
Type 2: ( a , a , a )
Clearly a can be any odd numbers from 1 , 3 , 5 , 7 , 9 , 1 1 i.e. we have 6 ways of choosing a and so there are 6 distinct ordered pairs of a , a , a , type
From cases above, required probability = 1 1 3 9 0 + 6 = 1 3 3 1 9 6
Moderator note:
Yes, we have to be careful to ensure that we got all of the possible cases, and didn't double count them.
Total permutations (total possible choices, n=11; amount of numbers to choose from, r=3) = n r = 1 3 3 1 .
For the first criteria, the product must be an odd number.
There are 11 sets in total, but sets 2,4,6,8,10 (even no) would be excluded. In the remaining 6 sets there are 6 odd numbers: 1,3,5,7,9,11. Therefore there are only 6 × 6 × 6 = 2 1 6 “odd number-product” sets.
For the second criteria, two or more numbers can be the same.
Similiarly, to be consistent and fulfill the first criteria, the even numbered sets are not counted. And hence in the first roll, there would be 6 such reps; in the second roll 5 such; in the third roll 34such, which accounts to 6 × 5 × 4 = 1 2 0 .
All in all, the probability that the rolls do not fulfil the criteria (ie does not produce an “even number product” and that the three rolls have at least one repeated number is 1 3 3 1 2 1 6 − 1 2 0 = 1 3 3 1 9 6
Hence, the answer, B − A = 1 2 3 5
Your calculations do not make sense. First, you subtracted the cases where at least two rolls are the same, to get 9 9 0 / 1 3 3 1 . (This correctly represents the rolls where all the numbers are different.) You then ADDED 1 − 1 1 1 5 / 1 3 3 1 = 2 1 6 / 1 3 3 1 , but you want rolls that satisfy a second restriction (namely that all the rolls are odd), so you need to subtract further cases, not add.
Instead, you can get the answer as follows: There are 6 3 = 2 1 6 cases where all the rolls are odd (each roll is 1, 3, 5, 7, 9, or 11). But you also want the cases where at least two rolls are the same, so you subtract the cases where all the rolls are different. If all the rolls are different, then there are 6 possibilities for the first roll, 5 possibilities for the second roll, then 4 possibilities for the third roll, for a total of 6 ⋅ 5 ⋅ 4 = 1 2 0 . This means there are 2 1 6 − 1 2 0 = 9 6 rolls where at least two rolls are the same, so the desired probability is 9 6 / 1 3 3 1 .
The answer can actually be derived:
Probability of it having repetitions (within the odd set):
1 3 3 1 6 × 1 0 × 3 + 6 = 1 3 3 1 1 8 6
But I am going to type the rest of these below to show that using Aaron's method, I still get the same answer:
The probability of getting it odd is 1 3 3 1 2 1 6
The probability above contains sets with repetitions and sets with no repetitions. We want only the sets with repetitions, so we have to minus the sets without repetitions:
Probability of it having repetitions (within the odd set):
1 3 3 1 1 8 6
The probability of it having no repetitions (within the odd set): 1 3 3 1 2 1 6 − 1 3 3 1 1 8 6 = 1 3 3 1 3 0
So the probability of it being odd and with repetition is:
1 3 3 1 2 1 6 − 1 3 3 1 3 0 = 1 3 3 1 1 8 6
and therefore the answer is B − A = 1 1 4 5 ?
(Still get the same answer)
See comment
Outcomes- { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 }
Total cases= ( 1 1 1 ) ( 1 1 1 ) ( 1 1 1 ) =1331
No. of cases in which the number on die are odd and distinct = ( 1 6 ) ( 1 5 ) ( 1 4 ) =120
Total cases in which numbers on die are odd(distinct or same) = ( 1 6 ) ( 1 6 ) ( 1 6 ) =216
Total cases in which atleast two numbers out of three which appear on die are same (Favourable)= 216-120=96
Probability of given event= T o t a l F a v o u r a b l e = 1 3 3 1 9 6
Hence, A=96 B=1331
Hence required answer is: 1331-96= 1235