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MATH1715 Tutee Revision Questions

MATH1715 Tutee Revision Questions

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Question: 26/12/2010<br />

The probability that a certain type of inoculation takes effect is 0.995. Using a Poisson<br />

approximation or otherwise, compute the probability that at most 2 out of 400 people given<br />

the inoculation will find that it has not taken effect.<br />

Response: Let X be the number for whom the inoculation does not take effect. For n = 400<br />

people, X ∼ Bin(n = 400, p = 0.005). What is the approximate distribution satisfied by<br />

X? The question requires P{X ≤ 2}.<br />

Question: 26/12/2010<br />

Three factories manufacture 20%, 30% and 50% of computer chips a company sells. If<br />

the fractions of defective chips are 0.4%, 0.3% and 0.2%, respectively, what fraction of the<br />

defective chips come from the third factory?<br />

Response: Let D denote defective, and A, B, C the source factory. Thus we are told, for<br />

example, P{A} = 0.20 and P{D|A} = 0.004.<br />

We require P{C|D}. Thus use Bayes’s theorem.<br />

P{C|D} =<br />

P{D|C} P{C}<br />

P{D|A} P{A} + · · · + P{D|C}P{C} .<br />

Question: 26/12/2010<br />

A box contains 3 blue balls and 5 red balls. Ann and Bob take turns (with Ann going first)<br />

to draw a ball from the box, without replacement, until a blue ball is drawn. What is the<br />

probability that Ann will draw the blue ball first?<br />

Response: Draw a tree diagram is one way. On the first stage Ann either draws blue (with<br />

probability 3/8) or red (with probability 5/8). If Ann draws red on stage one, on the second<br />

stage Bob either draws blue (with probability 3/7) or red (with probability 4/7). If Bob<br />

draws red on stage two, on the third stage Ann either draws blue or red, and so on. You<br />

require all routes which end with Ann drawing blue.<br />

The question can be answered in other ways. For example, let Ar denote Ann’s draw on<br />

stage r, r = 1, 3, 5, 7, and Bs denote Bob’s draw on stage s, s = 2, 4, 6, 8.<br />

P{Ann wins} = P{A1 = blue} + P{(A1 = red) ∩ (B2 = red) ∩ (A3 = blue)} + · · ·<br />

where P{A1 = blue} = 3,<br />

and 8<br />

and so on.<br />

P{(A1 = red) ∩ (B2 = red) ∩ (A3 = blue)} = 5 4 3<br />

× ×<br />

8 7 6 ,<br />

Question: 26/12/2010<br />

In the Euro millions lottery, a player must choose 5 correct digits from {1, 2, . . ., 50} on the<br />

main board and two “lucky stars” from the nine digits 1-9. The lottery draw yields a “5+2”<br />

12

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