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Worked solution

Packets, nested counts and a modal-parameter range

A manufacturer produces a large number of coloured glowing sticks, which are packed in packets of 8. Each packet consists of randomly chosen coloured glowing sticks. On average, 12% of the glowing sticks are green.

(i) Explain the significance of ‘large number’ in the first sentence.

(ii) Find the probability that a randomly chosen packet of glowing sticks contains at least 3 green glowing sticks.

(iii) Ian buys 40 randomly chosen packets of glowing sticks for a concert. Find the probability that fewer than 6 of these packets contain at least 3 green glowing sticks.

On average, the proportion of glowing sticks that are blue is pp. It is known that the modal number of blue glowing sticks in a packet is 3.

(iv) Use this information to find exactly the range of values that pp can take.

[2018/YJC/II/8]

CJC · Discussion · Question 9

Answer

(i) Approximately constant success probability. (ii) 0.06080.0608. (iii) 0.9670.967. (iv) 1/3<p<4/91/3<p<4/9 for a unique mode; include both endpoints if tied modes are allowed.

(i)

A large production pool means selecting a few sticks has negligible effect on the green proportion, so the success probability can be treated as approximately constant at 0.12. This supports the binomial approximation and approximately independent draws. The source states the constant-probability point.

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(ii)

Let X∼B(8,0.12)X\sim B(8,0.12) count green sticks in a packet. Then P(X≥3)=1−P(X≤2)=0.060789174888…=0.0608P(X\geq3)=1-P(X\leq2)=0.060789174888\ldots=0.0608 (3 s.f.).

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(iii)

Let Y count, out of 40 packets, those containing at least three green sticks. For independent packets, Y∼B(40,q)Y\sim B(40,q) with q equal to the probability in (ii). The source uses q=0.060789. Keeping full precision, P(Y<6)=P(Y≤5)=0.9673379941…=0.967P(Y<6)=P(Y\leq5)=0.9673379941\ldots=0.967 (3 s.f.).

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(iv)

Let W∼B(8,p)W\sim B(8,p) count blue sticks. Under the source’s unique-mode interpretation, require P(W=3)>P(W=2)P(W=3)>P(W=2) and P(W=3)>P(W=4)P(W=3)>P(W=4). The first inequality is 56p3(1−p)5>28p2(1−p)656p^3(1-p)^5>28p^2(1-p)^6, giving 2p>1−p and p>1/3. The second is 56p3(1−p)5>70p4(1−p)456p^3(1-p)^5>70p^4(1-p)^4, giving (4/5)(1−p)>p(4/5)(1-p)>p and p<4/9. Therefore 1/3<p<4/9. At p=1/3 the counts 2 and 3 tie; at p=4/9 the counts 3 and 4 tie. If the wording permits 3 merely to be one of the modes, both boundaries are included instead.

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Common mistakes

The source’s strict inequalities assume a unique mode. A tie is a separate interpretation, not a numerical rounding issue.

Solution notes

Source ambiguity in (iv): the prompt says “the modal number … is 3” without expressly excluding a tied mode. The source solution chooses strict neighbouring inequalities and the open interval. That answer is retained as the unique-mode interpretation; allowing 3 as a tied mode gives [1/3,4/9].

The printed definition of Y in (iii) omits the word “green” once; the question and the model clearly concern at least three green sticks. The reviewed explanation retains the intended event explicitly.

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