Probability - coin flip problem

[ENgez]

Homework Statement

you flip a fair coin (one which has equal odds of bieng HEADS or TAILS.
q = odds of getting TAILS=0.5 , p = odds of getting HEADS=0.5
A1 is the event on which the first two flips are HEADS.
A2 is the event on which the first two flips are HEADS and TAILS in that order.
A3 is the event on which the first flip is TAILS.
we mark X as the number of flips until we get two consecutive HEADS

a) calculate : $$P(X=k|Ai) i=1,2,3$$ if not directly calculatable express using $$p_{x}(k) and p_{x}(k-1), p_{x}(k-2)$$

b) calculate EX (expectancy of x)

Homework Equations

expectancy is for a descrete variable : EX=$$\sum k*P_{x}(k)$$

The Attempt at a Solution

first of all $$\ 2<k<\infty$$ because you need atleast two flips.
$$\ P(X=k|A1)=1$$

now this is where it gets a bit tricky for me:

$$\ P(X=k|A2)=p*q*(q+sqrt(p*q))^{k-4}$$

the p*q is becuase of the two first flips.
now we have k-4 flips remaining (we already "used up" two flips and we will "use up" another two flips to get the consecutive HEADS)

we define s=p*q. the k-4 remaining flips can be any combination of s and q so that there are exactly k-4 p's and q's, so s is actually 2 flips.

so all possible combinations are: $$\ k-4C1*s^{(k-3)/2}*q^{1} + k-4C2*s^{(k-2)/2}*q^{2} + ... = (q+sqrt(s))^{k-4}$$

therefore $$\ P(X=k|A2) = p*q*(p+sqrt(p*q))^{k-4}*p^{2} = 1/16$$

and by the same reasoning $$\ P(X=k|A2) = q*(p+sqrt(p*q))^{k-4}*p^{2} = 1/8$$

these results seem a little too simple. am i doing something wrong here?

I haven't tried b) yet.

 

[mighty2000]

Thank you for your post. Your calculations for P(X=k|A1) and P(X=k|A2) seem correct. However, P(X=k|A3) is not correct. It should be P(X=k|A3)=q(p+q)^{k-1}.

For part b), you can use the formula for the expectancy of a geometric distribution, which is EX=1/p. In this case, p=0.5, so EX=2 flips.

I hope this helps. Keep up the good work!