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Worked solution

Read a binomial mode and an expected number of qualifying samples

Samples, each of 12 articles, are taken at random from a large consignment in which 25% of articles are defective.

Find the number of defective articles which is most likely to occur in a single sample, and find the probability of obtaining this number.

If 100 samples of 12 articles are to be examined, calculate the number of samples in which you would expect to find 4 or more defective articles.

CJC · Review Problems · Question 12

Answer

Most likely count 33, probability 0.2580.258; expected number of samples 35.135.1.

Most likely count and its probability

Use X∼B(12,0.25)X\sim B(12,0.25), matching the question and printed distribution. The source prose says “out of 8”, a typo. The printed comparison table is:

xxP(X=x)P(X=x)
20.2323
30.2581
40.1936

The unique mode is 3, with probability P(X=3)≈0.2581036091=0.258P(X=3)\approx0.2581036091=0.258 (3 s.f.). As an authored check, (12+1)(.25)=3.25(12+1)(.25)=3.25 is nonintegral, so the binomial mode is its floor, 3.

How did this part go?

Source calculator probability table

The source screenshot enters binompdf(12,.25,X) and shows the following displayed probabilities, with its cursor at X=3. These values are restored as accessible text rather than reproduced as a calculator image:

xxDisplayed P(X=x)P(X=x)
00.03168
10.12671
20.23229
30.2581
40.19358
50.10324
60.04015
70.01147
80.00239
93.5×10−43.5\times10^{-4}
103.5×10−53.5\times10^{-5}

The explorer also computes the remaining possible counts 11 and 12, which are outside the visible screenshot rows.

How did this part go?

Expected number among 100 samples

The qualifying probability is q=P(X≥4)=1−P(X≤3)≈0.3512213826q=P(X\geq4)=1-P(X\leq3)\approx0.3512213826, printed as .35122 (5 s.f.). If YY counts qualifying samples among 100, the source uses Y∼B(100,q)Y\sim B(100,q) and E(Y)=100q≈35.12213826=35.1E(Y)=100q\approx35.12213826=35.1 (3 s.f.). An expectation need not be a whole number; independence between sample indicators is needed for the stated binomial model, but linearity of expectation alone suffices for this expectation when each has the same q.

How did this part go?

Common mistakes

An expected number such as 35.1 is not a claim that a realised sample count can be fractional.

Solution notes

No marks or paper attribution are printed. The calculator screenshot’s mathematical input and all visible rows are transcribed; its menu styling is not copied. Its “out of 8” prose typo is qualified.

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