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Worked solution

Correct the intermediate expansion before solving for the percentage

p% of students take Mathematics A and the remaining(100−p)% take Mathematics B.30% of Math A students fail and 40% of Math B students fail.

(i) Show that the probability exactly one of two students fails is 0.0002(2400−2p−0.01p2)0.0002(2400-2p-0.01p^2). marks

(ii) Given the probability both take Math A, conditional on exactly one failing, is 7/487/48, find p. marks

[2014/RI/II/9]

CJC · Review problems · Question 15

Answer

(i) 0.0002(2400−2p−0.01p2)0.0002(2400-2p-0.01p^2), with the source’s intermediate sign typo corrected; (ii) p=40p=40.

(i) Read both course/result branches

Let A,B denote the course and F failure. The course probabilities are p/100p/100 and (100−p)/100(100-p)/100; fail/pass probabilities are 0.3/0.7 and 0.4/0.6. Hence P(F)=0.3p/100+0.4(100−p)/100=(40−0.1p)/100P(F)=0.3p/100+0.4(100-p)/100=(40-0.1p)/100, and P(F′)=(60+0.1p)/100P(F')=(60+0.1p)/100.

How did this part go?

(i) Count the two independent-person orders

Using the source’s independent-student model, exactly one failure has probability

2!(40−0.1p100)(60+0.1p100)=210000(2400+4p−6p−0.01p2)=0.0002(2400−2p−0.01p2).\begin{aligned}&2!\left(\frac{40-0.1p}{100}\right)\left(\frac{60+0.1p}{100}\right)\\ & =\frac2{10000}(2400+4p-6p-0.01p^2)\\ & =0.0002(2400-2p-0.01p^2).\end{aligned}

The source’s intermediate line prints −4p+6p-4p+6p; those signs are reversed. The corrected expansion above yields the source’s intended final expression.

How did this part go?

(ii) Form the joint event in the conditional numerator

Both take Math A with exactly one failure means one Math-A failure and one Math-A pass, in two orders. Thus

(0.3p/100)(0.7p/100) 2!0.0002(2400−2p−0.01p2)=748,\frac{(0.3p/100)(0.7p/100)\,2!}{0.0002(2400-2p-0.01p^2)}=\frac7{48},

or 0.21p2/(2400−2p−0.01p2)=7/480.21p^2/(2400-2p-0.01p^2)=7/48.

How did this part go?

(ii) Solve and reject the inadmissible percentage

Cross-multiplication gives 10.08p2=16800−14p−0.07p210.08p^2=16800-14p-0.07p^2, hence 10.15p2+14p−16800=010.15p^2+14p-16800=0. The roots are p=40p=40 and p=−1200/29p=-1200/29. Since p is a percentage in[0,100], reject the negative root and retain 40.

How did this part go?

Compare your sketch

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Math A probability p/100, Math B probability (100 − p)/100. Given Math A: Fail 0.3, Pass 0.7. Given Math B: Fail 0.4, Pass 0.6. All source branches retained.
Math A probability p/100, Math B probability (100 − p)/100. Given Math A: Fail 0.3, Pass 0.7. Given Math B: Fail 0.4, Pass 0.6. All source branches retained.
Math A probability p/100, Math B probability (100 − p)/100. Given Math A: Fail 0.3, Pass 0.7. Given Math B: Fail 0.4, Pass 0.6. All source branches retained.

Common mistakes

The source’s printed intermediate cross terms have reversed signs. Keep p as a percentage and reject the negative root.

Solution notes

Both asks, complete source tree, all marginal/conditional steps, quadratic stages, both roots/rejection and marks retained. Printed −4 p+6 p typo corrected +4 p−6 p explicitly while original native text remains verbatim. Independent-student model used in source products is stated.

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