Question 1150678
.


            In my solution,  I systematically use the Excel function  ' BINOM.DIST(k, n, p, TRUE/FALSE) '.



(i)   Calculate the probability that at most  13  students will attend the stretching class.


<pre>
    P(<= 13 attend) = P(0) + P(1) + P(2) + . . . + P(13) = {{{sum(C(n,k)*0.95^(15-k)*0.05^k, k=0,k=13)}}} = 


                    = cumulative sum = BINOM.DIST(13, 15, 0.95, TRUE) = 0.1710.


    In this part, I confirm the answer by @Boreal.
</pre>


(ii) &nbsp;&nbsp;Given that at least &nbsp;12 &nbsp;students attend the stretching class, &nbsp;calculate the probability that there will be at most &nbsp;1 &nbsp;student absent.


<pre>
     "Given that at least &nbsp;12 &nbsp;students attend" means that the given part is

         P(12 attend) + P(13 attend) + P(14 attend) + P(15 attend).


     Next,  

         P(12 attend) = {{{C(12,15)*0.95^12*0.05^3}}} = BINOM.DIST(12, 15, 0.95, FALSE) = 0.030733;

         P(13 attend) = {{{C(13,15)*0.95^13*0.05^2}}} = BINOM.DIST(13, 15, 0.95, FALSE) = 0.134752;

         P(14 attend) = {{{C(14,15)*0.95^14*0.05^1}}} = BINOM.DIST(14, 15, 0.95, FALSE) = 0.365756;

         P(15 attend) = {{{C(15,15)*0.95^15*0.05^0}}} = BINOM.DIST(15, 15, 0.95, FALSE) = 0.463291.


     Now, they ask to calculate the conditional probability


         P = {{{(P(14) + P(15))/(P(12) + P(13) + P(14) + P(15))}}} = {{{(0.365756 + 0.463291)/(0.030733 + 0.134752 + 0.365756 + 0.463291)}}} = {{{0.829047/0.994533}}} = 0.833605.     <U>ANSWER</U>


     In this part, my answer is  <U>different</U>  from that of @Boreal.
</pre>