In a given week, it is estimated that the probability of at least one student becoming sick is 17/23. Students become sick independently from one week to the next. Find the probability that there are at least 3 weeks of no sick students before the 2nd week of at least one sick student.
Accepted Solution
A:
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]