Porisma

Worked solution

Translate birthday counts and find a strict minimum sample size

It may be assumed that the dates of birth in a large population are distributed throughout the year so that the probability of a randomly chosen person’s date of birth being in any particular month may be taken as 112\dfrac1{12}.

(i) Find the probability that, of 6 people chosen at random, exactly 2 will have birthdays in January. marks

(ii) Find the probability that, of 8 people chosen at random, at least one will have a birthday in January. marks

(iii) NN people are chosen at random. Find the least value of NN such that the probability that at least one will have a birthday in January exceeds 0.9. marks

(iv) Find the probability that, of 100 people chosen at random, at least 40 will have birthdays in May, June, July or Agusut. marks

[J84/I/12]

CJC · Review Problems · Question 9

Answer

(i) 0.07350.0735; (ii) 0.5010.501; (iii) least N=27N=27; (iv) 0.09660.0966.

(i)

Let X∼B(6,1/12)X\sim B(6,1/12). Then P(X=2)=(62)(1/12)2(11/12)4≈0.07354861915=0.0735P(X=2)=\binom62(1/12)^2(11/12)^4\approx0.07354861915=0.0735 to 3 significant figures.

How did this part go?

(ii)

The question specifies eight people, so use Y∼B(8,1/12)Y\sim B(8,1/12). Then P(Y≥1)=1−(11/12)8≈0.5014697533=0.501P(Y\geq1)=1-(11/12)^8\approx0.5014697533=0.501 to 3 significant figures. The source introductory line says nine, but its distribution and calculation correctly use eight.

How did this part go?

(iii)

For W∼B(N,1/12)W\sim B(N,1/12), require 1−(11/12)N>0.91-(11/12)^N>0.9. Thus (11/12)N<0.1(11/12)^N<0.1 and N>ln⁡(0.1)/ln⁡(11/12)≈26.5N>\ln(0.1)/\ln(11/12)\approx26.5. The least integer is N=27N=27. An authored minimality check gives about .895889 at N=26 and .904565 at N=27.

How did this part go?

(iv)

The four stated months have total probability 4/12=1/34/12=1/3 under the given model. Let U∼B(100,1/3)U\sim B(100,1/3). Then P(U≥40)=1−P(U≤39)≈0.09662307025=0.0966P(U\geq40)=1-P(U\leq39)\approx0.09662307025=0.0966 to 3 significant figures. The source switches the variable letter to Y in its final probability line; this is the event for U.

How did this part go?

Common mistakes

The uniform-month probability is supplied as a model assumption; use eight people in (ii), not the source prose typo.

Solution notes

The printed month spelling “Agusut” is retained in the prompt; it means August. The sample-size and variable-letter slips are preserved in raw extraction and corrected explicitly in the reviewed working.

Read the hint