Answer:
[tex]0.614[/tex]
Step-by-step explanation:
Let us suppose a time period "t" which is greater than 4 weeks.
Let us say
[tex]X = 1[/tex] for no sick student
[tex]X = 0[/tex] for sick student
Now the probability of at no sick student each week is given by
P [tex]= {1,1, 1,1}[/tex]
[tex](\frac{17}{23})^4[/tex]
There are other cases such as
[tex](1,1,1,0),(1,1,0,1),(1,0,1,1),(0,1,1,1)[/tex]
The probability of above cases is equal to
[tex](\frac{17}{23})^3*(1-\frac{17}{23})\\= (\frac{17}{23})^3 * \frac{6}{23}[/tex]
Now the probability of that there are at least 3 weeks of no sick students before the 2nd week of at least one sick student
= [tex](\frac{17}{23})^4 + 3*(\frac{17}{23})^3* \frac{6}{23} \\= \frac{171955}{279841} \\= 0.614[/tex]
Answer:
0.614
Step-by-step explanation:
Let the time be given by = t
and P(S ) = probability that a person is sick
P(s) = probability that a person is not sick
P(s) = [tex](\frac{17}{23})^{23}* (1-\frac{17}{23})\\[/tex]
Then the probability for that there are at least 3 weeks of no sick students before the 2nd week of at least one sick student is given by:
[tex](\frac{17}{23})(\frac{17}{23})(\frac{17}{23})(\frac{17}{23}) + \frac{6}{23} + 3(\frac{17}{23})^{3}\\ = 0.614[/tex]
