1. 6 different rolls =
6/6*5/6*4/6*3/6*2/6*1/6
A fair, 6-sided die is rolled 6 times independently. Assume that the results of the different rolls are independent. Let (a1,…,a6) denote a typical outcome, where each ai belongs to {1,…,6}.
Find the probability that the results of the 6 rolls are all different. (Answer with at least 3 decimal digits.)
For any outcome ω=(a1,…,a6), let R(ω) be the set {a1,…,a6}; this is the set of numbers that showed up at least once in the different rolls. For example, if ω=(2,2,5,2,3,5), then R(ω)={2,3,5}.
Find the probability that R(ω) has exactly two elements. (Answer with at least 3 decimal digits.)
Find the probability that R(ω) has exactly three elements.
7 answers
2. 155/7776
3. 25/108
3. 25/108
can you clarify the answer please
These are wrong
Solution of question 2:
(15*((2^6)-2))/(6^6)=0.0199
Hints: See this problem as a binary string. For example, fix a set {1,2}, then consider them as binary string {0,1}. So we know that 2^6 possible outcomes exist, which 2 out of them are {000000} and {111111}. Finally, (2^6)-2 outcomes can be possible for the set {1,2} as well as binary set {0,1}. As a result, 15 sets can be made each of which has (2^6)-2.
Solution of question 3:
20*(3^6-3*(2^6))=0.230.
Hints: It's the same as second problem. There are 3^6 possible outcomes for each three members-set which 3*(2^6) outcomes can't be considered for this problem. Thus, we can make 20 three members-sets each of which can be ordered in (3^6-3*(2^6)) ways.
Note that sample space is 6^6.
Best Regards, N-F.
(15*((2^6)-2))/(6^6)=0.0199
Hints: See this problem as a binary string. For example, fix a set {1,2}, then consider them as binary string {0,1}. So we know that 2^6 possible outcomes exist, which 2 out of them are {000000} and {111111}. Finally, (2^6)-2 outcomes can be possible for the set {1,2} as well as binary set {0,1}. As a result, 15 sets can be made each of which has (2^6)-2.
Solution of question 3:
20*(3^6-3*(2^6))=0.230.
Hints: It's the same as second problem. There are 3^6 possible outcomes for each three members-set which 3*(2^6) outcomes can't be considered for this problem. Thus, we can make 20 three members-sets each of which can be ordered in (3^6-3*(2^6)) ways.
Note that sample space is 6^6.
Best Regards, N-F.
Solution of question 3:
Thanks to Nima for the clarification. I think he forgot to take into account the one set consisting of 3's only, i.e {333333}, which is accounted for by the -1 in this formula. Th answer is practically the same:
20*(3^6-3*(2^6)-1)=0.2298.
Thanks to Nima for the clarification. I think he forgot to take into account the one set consisting of 3's only, i.e {333333}, which is accounted for by the -1 in this formula. Th answer is practically the same:
20*(3^6-3*(2^6)-1)=0.2298.
@Hamzeh
I think the right answer is:
20*(3^6-2*3*(2^6)-3)=1335/5832=0/2289
I think the right answer is:
20*(3^6-2*3*(2^6)-3)=1335/5832=0/2289
2 answers