F.3 Probability

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

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