# Joint probability distribution table .

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.

• "until either two heads or two tails have been tossed" you mean consecutive? – Tom Nov 22 '19 at 16:52
• no, they don't have to be consecutive @Tom – Morena Dragomir Nov 22 '19 at 16:53
• ok - so how many tosses will do you at maximum? – Tom Nov 22 '19 at 17:02
• 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. – Tom Nov 22 '19 at 17:25
• 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) – Tom Nov 22 '19 at 17:28

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 $$1$$.