Porisma

Worked solution

Solve a rejection probability and then model accepted cartons

In a carton of apples, a sample of 8 apples is taken and examined for spoilt apples.

(i) State, in context, an assumption for the number of spoilt apples in the sample to be modelled by a binomial distribution. mark

The number of spoilt apples in a random sample of size 8 may be modelled by the distribution B(8,p)B(8,p). If at least 2 apples in a sample are found to be spoilt, the carton is rejected. It is known that the probability of a carton being rejected is 0.04.

(ii) Write down an equation satisfied by pp and find the value of pp. marks

(iii) 60 cartons of apples are loaded onto a lorry. Find the probability that more than 56 cartons loaded onto the lorry are not rejected. marks

[2013/HCI/6(modified)]

CJC · Review Problems · Question 5

Answer

(ii) (1−p)8+8p(1−p)7=0.96(1-p)^8+8p(1-p)^7=0.96, p≈0.0410p\approx0.0410; (iii) 0.7810.781.

(i)

The source accepts either that each sampled apple has the same probability of being spoilt, or that whether one sampled apple is spoilt is independent of the others. The binomial model uses both constant probability and independence (approximately for a negligible without-replacement sample).

How did this part go?

(ii)

Let A∼B(8,p)A\sim B(8,p) count spoilt apples. Rejection means A≥2A\geq2, so P(A≥2)=0.04P(A\geq2)=0.04, equivalently P(A=0)+P(A=1)=0.96P(A=0)+P(A=1)=0.96. Thus (1−p)8+8p(1−p)7=0.96(1-p)^8+8p(1-p)^7=0.96. Solving on 0≤p≤10\leq p\leq1 gives p≈0.04104647418p\approx0.04104647418, or the printed p=0.0410p=0.0410 to 3 significant figures.

How did this part go?

(iii)

Let CC count cartons not rejected. Assuming independent carton decisions with the same acceptance probability, C∼B(60,0.96)C\sim B(60,0.96). More than 56 means C≥57C\geq57. Thus P(C>56)=1−P(C≤56)≈0.7813435261=0.781P(C>56)=1-P(C\leq56)\approx0.7813435261=0.781 to 3 significant figures. This uses the given carton rejection probability 0.04 directly, not a rejection probability recomputed from rounded p.

How did this part go?

Common mistakes

Keep the given 0.96 acceptance probability exact when moving to cartons.

Read the hint