deeper probability (1 Viewer)

[OLDMAN]

Method 2


Let p= P(Bob wins)

Then p= 1/4+1/2*1/4+1/2*1/4*p+1/4*p
where 1/4=P(Bob wins first toss)
1/2*1/4=P(Bob wins on second toss after a 1 Head toss)
1/2*1/4=P(TT after a 1 Head toss)
1/4=P(TT on first toss)
Solve for p to get 3/5.

Yes Archman, I could now see Affinity's sol.

 

[Affinity]

bah.. must be careful in the HSC then

 

[OLDMAN]

Method 3

P(Bob wins w/o revisiting start node)= 1/4+1/2*1/4=3/8
P(Penny wins w/o revisiting start node)= 1/4

P(Bob wins)=(3/8)/(3/8+1/4)
P(Penny wins=(1/4)/(3/8+1/4)

 

[adosh]

<HR style="COLOR: #d1d1e1" SIZE=1><!-- / icon and title --><!-- message -->hello everyone,,, im having a problem wtih permuts combins,,,,now geha answers a questions as such "in how many ways can 8 people be arranged into two sets of four for tennis??" : (8C4x4C4)/2!= 35,,,now i understand this but heres a similar fitzpatrick question

find the number of wats in which 6 women and 6 men can be arranged in three sets of four for tennis wihcout restrictions??? and the asnwer is 155925.....now why cant we apply gehas method to this questuion, that is

(12C4x8C4x4C4)/3!=5775 but this is way off from 155925,,,,couls someone please help..thanks in advance

 

[jkwii]

did u find the factor you are out by? its 27 or 3^3.

i.e. the above answer has taken into account the ways you can choose pairs - there are 3 ways to pair 4 people right?

this happen 3 times for 3 games of tennis so the answer is 3^3 x 5775 = 155925