標題:
probability and trigo
發問:
1. if 2 dice are thrown,a(i) what is the probability that the sum of the numers on the 2 dice is greater than 9a(ii) what is the probability that the sum of the numers on the 2 dice is greater than 9 or the no. on the two dice are equal.b. in a game, 2 dice are thrown. in each throw, 2points are gained if the... 顯示更多 1. if 2 dice are thrown, a(i) what is the probability that the sum of the numers on the 2 dice is greater than 9 a(ii) what is the probability that the sum of the numers on the 2 dice is greater than 9 or the no. on the two dice are equal. b. in a game, 2 dice are thrown. in each throw, 2points are gained if the sum of the no. on the 2 dice is greater than 9 or the no. on the 2dice are equal; otherwise point is lost. Using the result in a ii , find the probability of b(i) losing a total of 2 points in two throws b(ii) gaining a total of 3 points in three throws 2. in any triangle ABC, prove that a. a=b cos C+ c cos B b. a^2 / ( b^2-c^2) = (sinA)^2 / (sin B)^2-(sin C)^2
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最佳解答:
(1)(a)(i) The sum is greater than 9, the probable combinations are: (4,6), (6,4), (5,5), (5,6), (6,5) and (6,6) Probability = 6/36 = 1/6 (a)(ii) Additional combinations for equal dice are (1,1), (2,2), (3,3) and (4,4) Total probability = (6 + 4)/36 = 5/18 (b)(i) Let P = "the sum of the no. on the 2 dice is greater than 9 or the no. on the 2 dice are equal" losing 2 points in 2 throws => both throws are not P Probability = (1 - 5/18)^2 = (13/18)^2 = 169/324 (ii) Gaining 3 points in 3 throw means two P's and one not P Probability = 3C1 * (5/18)^2 * (13/18) = 975/5832 = 325/1944 (2)(a) By cosine rule cos C = (a^2 + b^2 - c^2) / 2ab cos B = (a^2 + c^2 - b^2) / 2ac b cos C + c cos B = (a^2 + b^2 - c^2) / 2a + (a^2 + c^2 - b^2) / 2a = 2a^2 / 2a = a (b) By sine rule, sinA/a = sinB/b = sinC/c b = a sin B / sin A c = a sin C / sin A a^2 / (b^2 - c^2) = a^2 / [(a sin B / sin A)^2 - (a sin C / sin A)^2] = a^2 / {[a^2 (sinB)^2 - a^2 (sinC)^2]/(sinA)^2} = (sinA)^2 / [(sinB)^2 - (sinC)^2]
其他解答:63D0B758E2D502CC
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