Question 1150678: “Best Fitness” Company is going to hold a stretching class with 15 registered students. From the previous records of stretching classes, the average absence rate per student is 5%.
(i) Calculate the probability that at most 13 students will attend the stretching class.
(ii) Given that at least 12 students attend the stretching class, calculate the probability that there will be at most 1 student absent.
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 0.05 probability absent and 0.95 probability present
probability at most is 13 is 1- probably (14 or 15) are there
that second term is 15C14*0.95^14*0.05=0.3658 plus 15C15*0.95^15=0.4633
They add to 0.8290 (rounding at end)
probability overall of 13 or fewer is 1-0.8290=0.1710 ANSWER
probability of 12 attending is 15C12*0.95^12*0.05^3=0.0307
probability of 13 attending is 0.1348
0.0307
0.1348
0.3658
0.4633
Given that 12 attend, what is the probability it will be 14 or 15, is another way to phrase b.
That would be their sum over the sum of p(13,14,15)
0.8290/0.9638=0.8601
Answer by ikleyn(53937)  (Show Source):
You can put this solution on YOUR website! .
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.
P(<= 13 attend) = P(0) + P(1) + P(2) + . . . + P(13) =  =
= cumulative sum = BINOM.DIST(13, 15, 0.95, TRUE) = 0.1710.
In this part, I confirm the answer by @Boreal.
(ii) Given that at least 12 students attend the stretching class, calculate the probability that there will be at most 1 student absent.
"Given that at least 12 students attend" means that the given part is
P(12 attend) + P(13 attend) + P(14 attend) + P(15 attend).
Next,
P(12 attend) =  = BINOM.DIST(12, 15, 0.95, FALSE) = 0.030733;
P(13 attend) =  = BINOM.DIST(13, 15, 0.95, FALSE) = 0.134752;
P(14 attend) =  = BINOM.DIST(14, 15, 0.95, FALSE) = 0.365756;
P(15 attend) =  = BINOM.DIST(15, 15, 0.95, FALSE) = 0.463291.
Now, they ask to calculate the conditional probability
P =  =  =  = 0.833605. ANSWER
In this part, my answer is different from that of @Boreal.
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