probability problem

[dionys]

Hi..

I have an exercise and i want to know if i solved it correct.Can you help?

Three people A,B,C throw a coin (First is A,second comes B and finally C)
The first that will bring head wins and the game stops.

The exercise wants:
a)The sample space
b)Determinate the following events in the sample space
i)A={A WINS}
ii)B={B WINS)
iii)(A union B)complement
H:HEAD T:TAIL
a)sample space{H,TH,TTH,TTT}
bi)P(A)={H}=1/2
bii)P(B)={TH}=P(Acomplement intersection B)=1/4
biii)(A union B)complement=Acomplement intersection Bcomplement=
{TTH,TTH}=1/4

is it corrent?

[rajesh]

a) I think the sample space has countable infinite number of elements.
what if all A, B, C fail in their first trials? They again start with A. So ur sample space is extended to TTTH, TTTTH......etc,

bi)A can win in the first round with prob 1/2
or in the 2nd round with prob 1/16
or in the 3rd round with prob 1/128
.
.
.

Therefore P(A) = 1/2+1/16+1/128+....... = 4/7

bii) Smilarly P(B) = 1/4+1/32+.......= 2/7
biii)A and B r mutually exclusive. P(A union B) = P(A) + P(B) = 6/7

P(A union B)complement = 1-6/7 = 1/7.

Aliter: P(A union B) means either A wins or B wins. So P(A union B)complement means neither A nor B wins ==> P(C wins) = 1/7(find out as above).

[dionys]

You have right.
But can you explain me how did you find
"or in the 2nd round with prob 1/16
or in the 3rd round with prob 1/128"

i know that is:1/2*1/2*1/2*1/2

but can you explain me why we multiply the probablilities plz?

[rajesh]

because they r independent. that means whether B gets a H/T doesnt depend on the previous outcome(A' toss). similarly with C and so on........