Challenging probability question (1 Viewer)

[Bellow]

A fair coin is to be tossed repeatedly. For integers r and s, not both zero, let P(r,s) be the probability that a total of r heads are tossed before a total of s tails are tossedso that P(0,1) = 1 and P(1,0) = 0.

(i) Explain why, for r, s≥1,
P(r,s)=½P(r-1,s)+½P(r,s-1).
(ii)Find P(2,3) by using part (i).
(iii)By using induction on n=r+s-1, or otherwise, prove that
P(r,s)=[1/2ⁿ]{(nC0)+(nC1)+.....+(nC(s-1))}
for s≥1.

 

[haboozin]

A fair coin is to be tossed repeatedly. For integers r and s, not both zero, let P(r,s) be the probability that a total of r heads are tossed before a total of s tails are tossedso that P(0,1) = 1 and P(1,0) = 0.

(i) Explain why, for r, s≥1,
P(r,s)=½P(r-1,s)+½P(r,s-1).

first one needs a head more second one needs a tail more

so P(r,s) = p(H).P(r-1,s) + P(T)P(r,s-1)
p(h) = 1/2

p(t) = 1/2

so P(r,s)=½P(r-1,s)+½P(r,s-1).


(ii)Find P(2,3) by using part (i).

from i P(2,3)= 1/2p(1/3) + 1/2P(2,2)
P(2,2) = P(HH, HTH, THH) = 1/4 + 1/8 + 1/8 = 1/2

P(1,3)= (H, TH, TTH) = 1/2 + 1/4 + 1/8 = 7/8

sp P(2,3) = 1/2x7/8 + 1/2x1/2 = 7/16 + 1/4 = 11/16

(iii)By using induction on n=r+s-1, or otherwise, prove that
P(r,s)=[1/2ⁿ]{(nC0)+(nC1)+.....+(nC(s-1))}
for s≥1
i have no idea....

 

[Condog]

the induction one is incredibly hard but doable,
first u must consider the case for n=0 ie s=1 r=0 and use i) and the formula provided
then prove for n=1 in both cases when s=2 and r=0 and when s=1 and r=1 using same principle
then assume for n=k
then prove for n=k+1 in two cases
when s=k+2 and r=0
and when s >= 1 and r>=1 using i) for this part very complicated and requires a little bit of everythin

 

[justchillin]

That is straight from HSC 2000...induction isnt that hard either...I have a crafty way, i'll give it to u after english...

 

[KFunk]

Given that you're online at the moment I thought I'd ask, what's your crafty method?

 

[justchillin]

I'll wait till after monday...dont want to give too much away..especially now

 

[香港!]

Bellow, I think I have the solution to that question... if you want tell me

 

[ngai]

A fair coin is to be tossed repeatedly. For integers r and s, not both zero, let P(r,s) be the probability that a total of r heads are tossed before a total of s tails are tossedso that P(0,1) = 1 and P(1,0) = 0.

(i) Explain why, for r, s≥1,
P(r,s)=½P(r-1,s)+½P(r,s-1).
(ii)Find P(2,3) by using part (i).
(iii)By using induction on n=r+s-1, or otherwise, prove that
P(r,s)=[1/2ⁿ]{(nC0)+(nC1)+.....+(nC(s-1))}
for s≥1.

(iii) P(r,s) = P(throw r heads before s tails)
"throw r heads before s tails"...
this is equivalent to:
"in the first r+s-1 throws, he throws at least r heads", since:
- if he doesnt, then he has thrown <= r-1 heads, and so he has thrown >= s tails, so not r heads before s tails
- if he does, then he has thrown >= r heads and <= s-1 tails, so he will throw r heads before s tails (even if he takes many more throws to get the s tails)

and of course, P(in the first r+s-1 throws, he throws >= r heads) is the same as:
P(in the first r+s-1 throws, he throws <= s-1 tails)

then just divide into cases:
P(throws 0 tails) = nC0/2^n (of the n throws, choose 0 to be tails, then prob of a head or tail are both 1/2)
P(throws 1 tail) = ...
in fact, P(throws k tails) = nCk/2^n
so P(throws <= s-1 tails) = (1/2^n)[nC0 + nC1 + ... + nC(s-1)]

 

[who_loves_maths]

and there's nothing particularly difficult about the induction method either...
one just has to recall the pattern with Pascal's Triangle where two consecutive members of a line add to give the "sandwiched" member below them in the next line. {i think we learn this in... year 9}

i.e. mathematically: (nCk) + (nC[k+1]) = ([n+1]C[k+1])

knowing that, the other steps in the induction process is intuitively straightforward is it not? ...

 

[jmromeo]

Hard probability question

Well... this, definitively, is not an easy question. In fact the percentage of students that attempted this question in 2000 was very low. So, do not feel bad if you cannot do it, however it's nice to see how difficult problems can be solved.
I hope nothing like it is on the comming test.

 

[Sepulchres]

I'll wait till after monday...dont want to give too much away..especially now

Whoa, whatadickhead.

I dont really care about your "crafty" method but mate, that attitude wont take you far in life.

"What goes, around comes around, its Karma baby."