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Probability Questions
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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) -------------------------------------------------------------------------------------------- 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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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... 顯示更多 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.最佳解答:
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) -------------------------------------------------------------------------------------------- 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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