A biased die is rolled n times. The probabilities for 1,2,3 are 0.3 each. The probabilities for the other numbers are also identical to each other. So the probability for 4,5,6 each would be 0.033. Right ?
What is the probability to observe 2 times 1, 1 time 5, 2 times 4 and 1 time 3 ? If there is no mention about probability to observe 1 and 2 on dice, can I still use multinomial distribution probability formula for this problem ?
Your probability for 4,5, and 6 seem to be correct. If we assume that each roll of the die is independent, you should be able to multiply the probabilities of landing on each number of interest.
Pr(1)*Pr(1)*Pr(5)*Pr(4)*Pr(4)*Pr(3) = 0.3*0.3*0.033*0.033*0.033*0.3 = 1x10-6
I'm not exactly sure what you mean by "no mention about probability to observe 1 and 2"
"What is the probability to observe 3 times 6, 1 time 5, 2 times 4 and 1 time 3 ?" So here they mention about observing 6,5,4,3 on dice but do not mention about observing 1 and 2 (Faces of dice). So can we assume that their observance was 0 ?
If there is no mention about observing 1 and 2 can we still use multinomial distribution formula to get the probability ?
I assume that "What is the probability to observe 2 times 1, 1 time 5, 2 times 4 and 1 time 3" should be read as "What is the probability to observe 2 times 1, 1 time 5, 2 times 4 and 1 time 3, *when rolling the die 6 times*". With that essential addition, you do have information on the number of times you roll 1 or 2: zero. And then you can apply the multinominal distribution straightforwardly.
Also, when you want to have "What is the probability to observe 2 times 1, 1 time 5, 2 times 4 and 1 time 3, *when rolling the die k times*", with k>6, you can still apply the multinomial. You now know that you have k-6 1's or 2's. Because you do not need to distinguish between them, just 'pool' them: P(1 or 2) = 0.6, and again apply the multinomial.
- Badges
- Science topic
- Similar topics
- Mathematics
- Statistics
- Probability