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標題:
F.3 Probability
發問:
A man is going to toss three $5 coins. If he is allowed to take away those coins with the head showing up, the expected amount of coins that he can get is ? Step 唔該.
最佳解答:
其他解答:
P(H H H) = (1/2)^3 = 1/8P(2H 1T) = (3C1)(1/2)^3 = 3/8P(1H 2T) = (3C1)(1/2)^3 = 3/8P(T T T) = (1/2)^3 = 1/8The expected amount of coins that he can get = 5 [3(1/8) + 2(3/8) + 1(3/8) + 0(1/8)]= $ 7.5 2010-11-18 19:41:51 補充： If you haven't learnt 3C1 , please see the below. P(2H 1T) = P(H H T) + P(H T H) + P(T H H) = (1/2)^3 + (1/2)^3 + (1/2)^3 = 3(1/2)^3 = 3/8
F.3 Probability
發問:
A man is going to toss three $5 coins. If he is allowed to take away those coins with the head showing up, the expected amount of coins that he can get is ? Step 唔該.
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
You may not be able to understand the answers offered by the helper 雨後陽光 there for you may not learn Binomials if you were a S.3 student or you do not take Module 1. I provide an alternative here and see if you could understand. P ( 1H ) = P( 2T 1H ) = 3 P ( 1st T, 2nd T, 3rd H ) [ you need to care the ordering ] = P( TTH ) + P( THT ) + P( HTT ) = 3 ( 1/2 )^3 = 3/8 -------- (1) P ( 2H ) = P( 1T 2H ) = 3 P ( 1st T, 2nd H, 3rd H ) [ you need to care the ordering ] = P( THH ) + P( HHT ) + P( HTH ) = 3 ( 1/2 ) ( 1/2 )^2= 3/8 -------- (2) P ( 3H ) = P ( 1st H, 2nd H, 3rd H )= P( HHH ) = ( 1/2 )^3= 1/8 -------- (3) Expecting value = Exp. Value from 1H + Exp. Value from 2H + Exp. Value from 3H = $5 ( 3/8 ) + $ 10 ( 3/8 ) + $ 15 ( 1/8 ) = $ ( 15 + 30 + 15 )/8 = $ 60/8 = $ 7 1/2其他解答:
P(H H H) = (1/2)^3 = 1/8P(2H 1T) = (3C1)(1/2)^3 = 3/8P(1H 2T) = (3C1)(1/2)^3 = 3/8P(T T T) = (1/2)^3 = 1/8The expected amount of coins that he can get = 5 [3(1/8) + 2(3/8) + 1(3/8) + 0(1/8)]= $ 7.5 2010-11-18 19:41:51 補充： If you haven't learnt 3C1 , please see the below. P(2H 1T) = P(H H T) + P(H T H) + P(T H H) = (1/2)^3 + (1/2)^3 + (1/2)^3 = 3(1/2)^3 = 3/8
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