Probability Questions

[匿名]

2011-10-31 8:54 am

Over the course of a day in the foreign exchange markets, the values of the pound, the euro and the US dollar change relative to one another. Let A be the event that the value of the pound goes up against the dollar (i.e., that £1 buys more dollars at the end of the day than at the beginning), B the event that the value of the dollar goes up against that of the euro, C the event that the value of the pound goes up compared to the euro.Describe in terms of exchange rate movements the event (A ∪ B)c ∩ C. Explain why (A ∪ B)c and C are mutually exclusive events. Suppose that P(A) = P(B) = P(C) = 0.5 and that P(A ∩ B) = p, P(A ∩ C) = P(B ∩ C) = q. Derive an expression for q in terms of p. Draw a Venn diagram to illustrate the events and mark in the probabilities on the diagram. Is there a value of p which ensures that A and B are independent events, A and C are independent events, and B and C are independent events? Give a reason for your answer.

回答 (1)

Kevin

2011-11-10 8:35 am

✔ 最佳答案

P(event) denotes the probability of an event.

In the beginning, there are 3 coins with 4 heads and 2 tails.

i)

P(getting a head in 1st trial) = 4/6 = 2/3

In this case 1st coin chosen may be either an HH or an HT,

P(getting a head in 2nd trial, an HH is chosen as 1st coin) = (1/3)(1) = 1/3

P(getting a tail in 2nd trial, an HT is chosen as 1st coin) = (2/3)(1/2) = 1/3

∴P(getting a tail in 2nd trial, in general) = 1/3 + 1/3 = 2/3

∴P(getting two heads in two trails)

= P(getting a head in 1st trial) × P(getting a tail in 2nd trial, in general)

= 2/3 × 2/3

= 4/9 .......... (ans)

ii)

P(getting a tail in 1st trial) = 2/6 = 1/3

In this case 1st coin chosen must be a HT,

P(getting a head in 2nd trial) = 3/4

P(getting a tail in 2nd trial) = 1/4

∴P(getting two tails in two trails)

= P(getting a tail in 1st trial) × P(getting a tail in 2nd trial)

= 1/3 × 1/4

= 1/12 .......... (ans)

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We can also deduce

from i)

P(getting a head and a tail in two trails) = (2/3)[(1/3)(0) + (2/3)(1/2)] = 2/9

from ii)

P(getting a tail and a head in two trails) = (1/3)(3/4) = 1/4

Sum up probabilities of all 4 cases, we can get 4/9 + 1/12 + 2/9 + 1/4 = 1

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