Porisma

Worked solution

Convert weeks to trials and use a variance restriction

According to the school rules, a student who arrives at school after 0730 hours is considered late.

(i) During the school term, a boy cycles to school on 5 days each week. On any given day, the probability that he arrives after 0730 hours is 0.1. For a period of 4 weeks, calculate the probability that he is late on at least one day but not more than 3 days. marks

When the boy has cycled to school for the nnth time, the probability that he arrived at school late at least once is greater than 0.99. Calculate the least value of nn. marks

(ii) A girl travels to school by bus on 5 days a week. Over a long period of time, the variance of the number of days per week on which she is late is 0.8. Given that pp denotes the probability that she arrives late and that p<0.5p<0.5, evaluate pp. marks

[2007/SRJC/II/10(a)]

CJC · Review Problems · Question 11

Answer

(i) 0.7450.745 and least n=44n=44; (ii) p=0.2p=0.2.

(i) Four-week count

Four five-day weeks give 20 trials. Let X∼B(20,0.1)X\sim B(20,0.1) under the source independent-day model. Then P(1≤X≤3)=P(X≤3)−P(X=0)≈0.7454700220=0.745P(1\leq X\leq3)=P(X\leq3)-P(X=0)\approx0.7454700220=0.745 (3 s.f.).

How did this part go?

(i) Least number of journeys

Let W∼B(n,0.1)W\sim B(n,0.1). Require 1−0.9n>0.991-0.9^n>0.99, so 0.9n<0.010.9^n<0.01 and n>ln⁡(0.01)/ln⁡(0.9)≈43.70869n>\ln(0.01)/\ln(0.9)\approx43.70869. Hence least n=44. As a minimality check, n=43 gives about .989225 and n=44 about .990302. The source changes the variable letter to X in intermediate lines; the calculation is for W.

How did this part go?

(ii)

For the weekly count Y∼B(5,p)Y\sim B(5,p), variance 5p(1−p)=0.85p(1-p)=0.8. Thus p2−p+0.16=0p^2-p+0.16=0, with roots .8 and .2. The given p<0.5p<0.5 rejects .8, leaving p=0.2p=0.2.

How did this part go?

Common mistakes

The variance equation has two roots; the printed inequality selects the smaller one.

Solution notes

The source binomial calculations assume independent daily outcomes with the stated constant probabilities.

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