probability and trigo

2009-12-10 6:35 am

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

回答 (1)

用戶9112

2009-12-10 7:04 am

✔ 最佳答案

(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]

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