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A biased coin has probability $\frac{1}{4}$ of landing on ‘heads’ and $\frac{3}{4}$ of landing on ‘tails’ when tossed. The coin is tossed repeatedly until either two heads or two tails have been tossed. Let $X$ denote the total number of heads achieved and $Y$ the total number of tails achieved in the sequence of tosses. Thus, for example, if the first toss is a tail, and the second and third tosses are both heads then $X = 2$ and $Y = 1$.

(a) Describe the joint distribution of $X$ and $Y$ by a clearly labelled table.

(b) Find the marginal distributions of $X$ and $Y$ .

(c) Suppose that $Y = 2$. Find $P(X = x | Y = 2)$ for each possible value of $x$

I have tried to compute the table but I am not getting that the sum of probabilities add up to 1. I am sure I can do the second parts if I get the table right though.

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  • $\begingroup$ "until either two heads or two tails have been tossed" you mean consecutive? $\endgroup$ – Tom Nov 22 '19 at 16:52
  • $\begingroup$ no, they don't have to be consecutive @Tom $\endgroup$ – Morena Dragomir Nov 22 '19 at 16:53
  • $\begingroup$ ok - so how many tosses will do you at maximum? $\endgroup$ – Tom Nov 22 '19 at 17:02
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    $\begingroup$ You could think of it like: if i got HH, i still toss again, but the third toss is completely irrelevant as far as my X and Y are concerned. So for example X(HHT)=2, Y(HHT)=0 and X(HHH)=2, Y(HHH)=0. $\endgroup$ – Tom Nov 22 '19 at 17:25
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    $\begingroup$ Otherwise you could take as sample space {HH, HTH, HTT, THH, THT, TT} (sorry i misread your comment above where you took {HH, TT, HTH,THT} which would be wrong) $\endgroup$ – Tom Nov 22 '19 at 17:28
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The set of outcomes from this activity are $\{HH, HTH, HTT,TT, THT, THH\}$. The maximum number of coin tosses are three and hence the sample space is

$$\{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT\}$$

From the sample space , you know that $X \in \{0,1,2,3\}$ and $Y \in \{0,1,2,3\}$. We can now derive the joint probabilities. The sum of all values in the table will be $1$.

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