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Worked solution

Count faults across two boxes rather than test each box separately

The probability that a particular make of light bulb is faulty is 0.2. The light bulbs are packed in boxes of 12.

(a) Find the probability that in a certain box, there are

(i) no faulty light bulbs

(ii) more than 3 faulty light bulbs

(b) A buyer accepts a consignment of 50 boxes, if, when he chooses two boxes at random, he finds that they contain no more than one faulty light bulb altogether. Find the probability that he will accept the consignment.

CJC · Review Problems · Question 4

Answer

(a)(i) 0.06870.0687; (a)(ii) 0.2050.205; (b) 0.03310.0331.

(a)(i)

Let XX count faulty bulbs in a box of 12. Under independent faults with common probability 0.2, X∼B(12,0.2)X\sim B(12,0.2). Then P(X=0)=0.812≈0.06871947674=0.0687P(X=0)=0.8^{12}\approx0.06871947674=0.0687 to 3 significant figures.

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(a)(ii)

More than three means at least four, so P(X>3)=1−P(X≤3)≈0.2054310502=0.205P(X>3)=1-P(X\leq3)\approx0.2054310502=0.205 to 3 significant figures.

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(b) Method 1: two box counts

For independent box counts X1,X2X_1,X_2, acceptance is the disjoint union (0,0),(0,1),(1,0)(0,0),(0,1),(1,0). Thus P(accept)=P(X=0)2+2P(X=0)P(X=1)≈0.03305656538=0.0331P(\text{accept})=P(X=0)^2+2P(X=0)P(X=1)\approx0.03305656538=0.0331 to 3 significant figures. The source intermediate product line mistakenly ends with P(X1=1)P(X2=1)P(X_1=1)P(X_2=1); it should be P(X1=1)P(X2=0)P(X_1=1)P(X_2=0). Its following simplified expression and final answer are correct.

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(b) Method 2: all sampled bulbs

Two boxes contain 24 bulbs. Let Y∼B(24,0.2)Y\sim B(24,0.2) count their faults; acceptance requires Y≤1Y\leq1. Hence P(Y≤1)=0.824+24(0.2)(0.8)23≈0.03305656538=0.0331P(Y\leq1)=0.8^{24}+24(0.2)(0.8)^{23}\approx0.03305656538=0.0331. The consignment has 50 boxes, but only the 24 tested bulbs enter this acceptance event.

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Common mistakes

Allowing one fault in each box would allow two altogether and change the acceptance rule.

Solution notes

No paper attribution or marks are printed for this question. The source two-box product typo is retained in the raw transcription and corrected explicitly in the reviewed explanation. The model assumes independent bulbs with common fault probability.

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