2011 A-level H2 Mathematics (9740) Paper 2 Question 9 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
(a)
P(faulty)
= P(made by A and faulty) + P(made by B and faulty)
= 0.6(0.05) + 0.4(0.07)
= 0.058

(b)
P(made by A | faulty)
= \frac{\mathrm{P(made~by~A~and~faulty)}}{\mathrm{P(faulty)}}
= \frac{0.6(0.05)}{0.058}
= \frac{15}{29}

(ii)
(a)
P(exactly one of them is faulty)
= 0.058 \times (1 - 0.058) \times 2!
= 0.109272 (exact)

(b)
P(both were made by A | exactly one is faulty)
= \frac{\mathrm{P(both~were~made~by~A~and~exactly~one~is~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}
= \frac{\mathrm{P(one~is~made~by~A~and~faulty,~the~other~is~made~by~A~and~not~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}
= \frac{0.6(0.05) \times 0.6(0.95) \times 2!}{0.109272}
= \frac{1425}{4553}

KS Comments:

Question can be easily solved by drawing a tree diagram. Do take note that we only wrong off NON exact answers to 3sf, so for (iia), we keep the full exact answer.

Leave a Comment

Contact Us

CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

Not readable? Change text. captcha txt

Start typing and press Enter to search