(ECE302 Problem - Zihui Liu) |
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| − | [[Category: | + | [[Category:ECE302Spring2013Boutin]] [[Category:ECE]] [[Category:ECE302]] [[Category:probability]] [[Category|problem solving]] |
| + | [[Category:independence]] | ||
| − | + | Problem: | |
| + | In a chess tournament, the probability of A wins B is 0.4. And the probability of C wins D is 0.5. The probability of C wins A is 0.6. The rule of tournament is A vs B in the first round. The second round is the person who wins in the first round vs C. The third round is the person who wins in the second round vs the person who lose in the first round. The fourth round is the winner in the third round vs the the person who lose in the second round. And the question is: | ||
| + | (a) Calculate the probability of B wins four round continously. | ||
| + | (b) Calculate the probablity of C wins three round continously. | ||
| − | + | Answer: | |
| + | (a)If B wants to win four round continously, | ||
| + | The first round is A vs B. And B wins | ||
| + | The first round is B vs C. And B wins | ||
| + | The first round is B vs A. And B wins | ||
| + | The first round is B vs C. And B wins | ||
| + | Since the probability of each time B wins is independently. And we define it is event A | ||
| + | so P(A) = (1-0.4)^2 + 0.5^2 = 0.09 | ||
| + | So the probability of B wins four round continously is 0.09. | ||
| + | (b) If C wants to win three round continously, | ||
| + | The first round is A vs B. And A wins ,or B wins. So there are two seperately solution | ||
| − | [[ | + | (1) If A wins in the first round, |
| + | the second round is A vs C and C wins | ||
| + | the third round is C vs B and C wins | ||
| + | the fourth round is C va A and C wins | ||
| + | (2) If B wins in the first round | ||
| + | the second round is B vs C and C wins | ||
| + | the third round is C vs A and C wins | ||
| + | the fourth round is C va B and C wins | ||
| + | |||
| + | Since the probability of each person wins is independently. And we define it is event B | ||
| + | so P(B) = 0.4 * 0.6^2 * 0.5 + (1-0.4) * 0.5^2 * 0.6 = 0.162 | ||
| + | So the probablity of C wins three round continously is 0.162. | ||
| + | |||
| + | [[Bonus_point_1_ECE302_Spring2012_Boutin|Back to first bonus point opportunity, ECE302 Spring 2013]] | ||
| + | |||
| + | -------------------------------------------------------------------------------- | ||
Latest revision as of 18:15, 27 January 2013
problem solving
Problem: In a chess tournament, the probability of A wins B is 0.4. And the probability of C wins D is 0.5. The probability of C wins A is 0.6. The rule of tournament is A vs B in the first round. The second round is the person who wins in the first round vs C. The third round is the person who wins in the second round vs the person who lose in the first round. The fourth round is the winner in the third round vs the the person who lose in the second round. And the question is:
(a) Calculate the probability of B wins four round continously. (b) Calculate the probablity of C wins three round continously.
Answer:
(a)If B wants to win four round continously,
The first round is A vs B. And B wins The first round is B vs C. And B wins The first round is B vs A. And B wins The first round is B vs C. And B wins
Since the probability of each time B wins is independently. And we define it is event A so P(A) = (1-0.4)^2 + 0.5^2 = 0.09 So the probability of B wins four round continously is 0.09.
(b) If C wants to win three round continously,
The first round is A vs B. And A wins ,or B wins. So there are two seperately solution
(1) If A wins in the first round,
the second round is A vs C and C wins
the third round is C vs B and C wins
the fourth round is C va A and C wins
(2) If B wins in the first round
the second round is B vs C and C wins
the third round is C vs A and C wins
the fourth round is C va B and C wins
Since the probability of each person wins is independently. And we define it is event B so P(B) = 0.4 * 0.6^2 * 0.5 + (1-0.4) * 0.5^2 * 0.6 = 0.162 So the probablity of C wins three round continously is 0.162.
Back to first bonus point opportunity, ECE302 Spring 2013
