if p is the probability that some bin ends up with 3 balls and q is the probability that every bin ends up with 4 balls. pq is 16.
First, let us label the bins with 1,2,3,4,5.
Applying multinomial distribution with parameters n=20 and p1=p2=p3=p4=p5=15 we find that probability that bin1 ends up with 3, bin2 with 5 and bin3, bin4 and bin5 with 4 balls equals:
5−2020!3!5!4!4!4!
But of course, there are more possibilities for the same division (3,5,4,4,4) and to get the probability that one of the bins contains 3, another 5, et cetera we must multiply with the number of quintuples that has one 3, one 5, and three 4's. This leads to the following:
p=20×5−2020!3!5!4!4!4!
In a similar way we find:
q=1×5−2020!4!4!4!4!4!
So:
pq=20×4!4!4!4!4!3!5!4!4!4!=20×45=16
thus, pq = 16.
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Answer:
a. 1/2
b. 1 hair cut
c. 2.5
Step-by-step explanation:
4/8 divde by 4 to simplify and that gets you 1/2
8/4 is 2 so he does one hair cut every two hours
so you know how many he can do in 2 hours so add it up. It’s going to be 2 people in four hours so it will be half a hair cut in 5 hours
Given:
Minni is arranging 3 different music CDs in a row on a shelf.
To find:
The sample space for the arrangement of a jazz CD (J), a pop CD (P), and a rock CD (R).
Solution:
We know that the total number of ways to arrange n items is n!.
Minni is arranging 3 different music CDs in a row on a shelf. So, the total number of ways is:
The sample space for the arrangement of a jazz CD (J), a pop CD (P), and a rock CD (R) is:
{JPR, JRP, PJR, PRJ, RJP, RPJ}
Therefore, the required sample space is {JPR, JRP, PJR, PRJ, RJP, RPJ}.
Your answer is 160. Sorry about that answer above I laughed reading it hahaha.
Answer:
19
Step-by-step explanation:
