Tag Archives: Bayes’ Theorem

An Example of a Joint Distribution

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Probem 1
Let X be the value of one roll of a fair die. If the value of the die is x, we are given that Y \lvert X=x has a binomial distribution with n=x and p=\frac{1}{4} (we use the notation Y \lvert X=x \sim \text{binom}(x,\frac{1}{4})).

  1. Discuss how the joint probability function P[X=x,Y=y] is computed for x=1,2,3,4,5,6 and y=0,1, \cdots, x.
  2. Compute the conditional binomial distributions Y \lvert X=x where x=1,2,3,4,5,6.
  3. Compute the marginal probability function of Y and the mean and variance of Y.
  4. Compute P(X=x \lvert Y=y) for all applicable x and y.

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Discussion of Problem 1

Problem 2 is found at the end of the post.

Problem 1.1
This is an example of a joint distribution that is constructed from taking product of conditional distributions and a marginial distribution. The marginal distribution of X is a uniform distribution on the set \left\{1,2,3,4,5,6 \right\} (rolling a fiar die). Conditional of X=x, Y has a binomial distribution \text{binom}(x,\frac{1}{4}). The following is the sample space of the joint distribution of X and Y.

Figure 1

The joint probability function of X and Y may be written as:

\displaystyle (1) \ \ \ \ \ P(X=x,Y=y)=P(Y=y \lvert X=x) \times P(X=x)

Thus the probability at each point in Figure 1 is the product of P(X=x), which is \frac{1}{6}, with the conditional probability P(Y=y \lvert X=x), which is binomial. For example, the following diagram and equation demonstrate the calculation of P(X=4,Y=3)

Figure 2

\displaystyle \begin{aligned}(1a) \ \ \ \ \ P(X=4,Y=3)&=P(Y=3 \lvert X=4) \times P(X=4) \\&=\binom{4}{3} \biggl[\frac{1}{4}\biggr]^3 \biggl[\frac{3}{4}\biggr]^1 \times \frac{1}{6} \\&=\frac{12}{256} \end{aligned}

Problem 1.2
The following shows the calculation of the binomial distributions.

\displaystyle \begin{aligned} (2) \ \ \ Y \lvert X=1 \ \ \ \ \ &P(Y=0 \lvert X=1)=\frac{3}{4} \\&P(Y=1 \lvert X=1)=\frac{1}{4} \end{aligned}

\displaystyle \begin{aligned} (3) \ \ \ Y \lvert X=2 \ \ \ \ \ &P(Y=0 \lvert X=2)=\binom{2}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^2=\frac{9}{16} \\&P(Y=1 \lvert X=2)=\binom{2}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^1=\frac{6}{16} \\&P(Y=2 \lvert X=2)=\binom{2}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{16} \end{aligned}

\displaystyle \begin{aligned} (4) \ \ \ Y \lvert X=3 \ \ \ \ \ &P(Y=0 \lvert X=3)=\binom{3}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^3=\frac{27}{64} \\&P(Y=1 \lvert X=3)=\binom{3}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^2=\frac{27}{64} \\&P(Y=2 \lvert X=3)=\binom{3}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^1=\frac{9}{64} \\&P(Y=3 \lvert X=3)=\binom{3}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{64} \end{aligned}

\displaystyle \begin{aligned} (5) \ \ \ Y \lvert X=4 \ \ \ \ \ &P(Y=0 \lvert X=4)=\binom{4}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^4=\frac{81}{256} \\&P(Y=1 \lvert X=4)=\binom{4}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^3=\frac{108}{256} \\&P(Y=2 \lvert X=4)=\binom{4}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^2=\frac{54}{256} \\&P(Y=3 \lvert X=4)=\binom{4}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^1=\frac{12}{256} \\&P(Y=4 \lvert X=4)=\binom{4}{4} \biggl(\frac{1}{4}\biggr)^4 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{256} \end{aligned}

\displaystyle \begin{aligned} (6) \ \ \ Y \lvert X=5 \ \ \ \ \ &P(Y=0 \lvert X=5)=\binom{5}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^5=\frac{243}{1024} \\&P(Y=1 \lvert X=5)=\binom{5}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^4=\frac{405}{1024} \\&P(Y=2 \lvert X=5)=\binom{5}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^3=\frac{270}{1024} \\&P(Y=3 \lvert X=5)=\binom{5}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^2=\frac{90}{1024} \\&P(Y=4 \lvert X=5)=\binom{5}{4} \biggl(\frac{1}{4}\biggr)^4 \biggl(\frac{3}{4}\biggr)^1=\frac{15}{1024} \\&P(Y=5 \lvert X=5)=\binom{5}{5} \biggl(\frac{1}{4}\biggr)^5 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{1024} \end{aligned}

\displaystyle \begin{aligned} (7) \ \ \ Y \lvert X=6 \ \ \ \ \ &P(Y=0 \lvert X=6)=\binom{6}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^6=\frac{729}{4096} \\&P(Y=1 \lvert X=6)=\binom{6}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^5=\frac{1458}{4096} \\&P(Y=2 \lvert X=6)=\binom{6}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^4=\frac{1215}{4096} \\&P(Y=3 \lvert X=6)=\binom{6}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^3=\frac{540}{4096} \\&P(Y=4 \lvert X=6)=\binom{6}{4} \biggl(\frac{1}{4}\biggr)^4 \biggl(\frac{3}{4}\biggr)^2=\frac{135}{4096} \\&P(Y=5 \lvert X=6)=\binom{6}{5} \biggl(\frac{1}{4}\biggr)^5 \biggl(\frac{3}{4}\biggr)^1=\frac{18}{4096} \\&P(Y=6 \lvert X=6)=\binom{6}{6} \biggl(\frac{1}{4}\biggr)^6 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{4096} \end{aligned}

Problem 1.3
To find the marginal probability P(Y=y), we need to sum P(X=x,Y=y) over all x. For example, P(Y=2) is the sum of P(X=x,Y=2) for all x=2,3,4,5,6. See the following diagram

Figure 3

As indicated in (1), each P(X=x,Y=2) is the product of a conditional probability P(Y=y \lvert X=x) and P(X=x)=\frac{1}{6}. Thus the probability indicated in Figure 3 can be translated as:

\displaystyle \begin{aligned}(8) \ \ \ \ \ P(Y=2)&=\sum \limits_{x=2}^6 P(Y=2 \lvert X=x) P(X=x) \end{aligned}

We now begin the calculation.

\displaystyle \begin{aligned}(9) \ \ \ \ \ P(Y=0)&=\sum \limits_{x=1}^6 P(Y=0 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{3}{4}+\frac{9}{16}+\frac{27}{64} \\&+ \ \ \ \frac{81}{256}+\frac{243}{1024}+\frac{729}{4096} \biggr] \\&=\frac{10101}{24576} \end{aligned}

\displaystyle \begin{aligned}(10) \ \ \ \ \ P(Y=1)&=\sum \limits_{x=1}^6 P(Y=1 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{4}+\frac{6}{16}+\frac{27}{64} \\&+ \ \ \ \frac{108}{256}+\frac{405}{1024}+\frac{1458}{4096} \biggr] \\&=\frac{9094}{24576} \end{aligned}

\displaystyle \begin{aligned}(11) \ \ \ \ \ P(Y=2)&=\sum \limits_{x=2}^6 P(Y=2 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{16}+\frac{9}{64} \\&+ \ \ \ \frac{54}{256}+\frac{270}{1024}+\frac{1215}{4096} \biggr] \\&=\frac{3991}{24576} \end{aligned}

\displaystyle \begin{aligned}(12) \ \ \ \ \ P(Y=3)&=\sum \limits_{x=3}^6 P(Y=3 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{64} \\&+ \ \ \ \frac{12}{256}+\frac{90}{1024}+\frac{540}{4096} \biggr] \\&=\frac{1156}{24576} \end{aligned}

\displaystyle \begin{aligned}(13) \ \ \ \ \ P(Y=4)&=\sum \limits_{x=4}^6 P(Y=4 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{256}+\frac{15}{1024}+\frac{135}{4096} \biggr] \\&=\frac{211}{24576} \end{aligned}

\displaystyle \begin{aligned}(14) \ \ \ \ \ P(Y=5)&=\sum \limits_{x=5}^6 P(Y=5 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{1024}+\frac{18}{4096} \biggr] \\&=\frac{22}{24576} \end{aligned}

\displaystyle \begin{aligned}(15) \ \ \ \ \ P(Y=6)&=\sum \limits_{x=6}^6 P(Y=6 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{4096} \biggr] \\&=\frac{1}{24576} \end{aligned}

The following is the calculation of the mean and variance of Y.

\displaystyle \begin{aligned}(16) \ \ \ \ \ E(Y)&=\frac{10101}{24576} \times 0+\frac{9094}{24576} \times 1+\frac{3991}{24576} \times 2 \\&+ \ \ \ \ \frac{1156}{24576} \times 3+\frac{211}{24576} \times 4+\frac{22}{24576} \times 5 \\&+ \ \ \ \ \frac{1}{24576} \times 6 \\&=\frac{21504}{24576}\\&=0.875 \end{aligned}

\displaystyle \begin{aligned}(17) \ \ \ \ \ E(Y^2)&=\frac{10101}{24576} \times 0+\frac{9094}{24576} \times 1+\frac{3991}{24576} \times 2^2 \\&+ \ \ \ \ \frac{1156}{24576} \times 3^2+\frac{211}{24576} \times 4^2+\frac{22}{24576} \times 5^2 \\&+ \ \ \ \ \frac{1}{24576} \times 6^2 \\&=\frac{39424}{24576}\\&=\frac{77}{48} \end{aligned}

\displaystyle (18) \ \ \ \ \ Var(Y)=\frac{77}{48}-0.875^2=\frac{161}{192}=0.8385

Problem 1.4
The conditional probability P(Y=y \lvert X=x) is easy to compute since it is a given that Y is a binomial variable conditional on a value of X. Now we want to find the backward probability P(X= x \lvert Y=y). Given the binomial observation is Y=y, what is the probability that the roll of the die is X=x? This is an application of the Bayes’ theorem. We can start by looking at Figure 3 once more.

Consider P(X=x \lvert Y=2). In calculating this conditional probability, we only consider the 5 sample points encircled in Figure 3 and disregard all the other points. These 5 points become a new sample space if you will (this is the essence of conditional probability and conditional distribution). The sum of the joint probability P(X=x,Y=y) for these 5 points is P(Y=2), calculated in the previous step. The conditional probability P(X=x \lvert Y=2) is simply the probability of one of these 5 points as a fraction of the total probability P(Y=2). Thus we have:

\displaystyle \begin{aligned}(19) \ \ \ \ \ P(X=x \lvert Y=2)&=\frac{P(X=x,Y=2)}{P(Y=2)} \end{aligned}

We do not have to evaluate the components that go into (19). As a practical matter, to find P(X=x \lvert Y=2) is to take each of 5 probabilities shown in (11) and evaluate it as a fraction of the total probability P(Y=2). Thus we have:

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 2 \bold )
\displaystyle \begin{aligned}(20a) \ \ \ \ \ P(X=2 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{16}}{\displaystyle \frac{3991}{24576}} =\frac{256}{3991} \end{aligned}

\displaystyle \begin{aligned}(20b) \ \ \ \ \ P(X=3 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{9}{64}}{\displaystyle \frac{3991}{24576}} =\frac{576}{3991} \end{aligned}

\displaystyle \begin{aligned}(20c) \ \ \ \ \ P(X=4 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{54}{256}}{\displaystyle \frac{3991}{24576}} =\frac{864}{3991} \end{aligned}

\displaystyle \begin{aligned}(20d) \ \ \ \ \ P(X=5 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{270}{1024}}{\displaystyle \frac{3991}{24576}} =\frac{1080}{3991} \end{aligned}

\displaystyle \begin{aligned}(20e) \ \ \ \ \ P(X=6 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{1215}{4096}}{\displaystyle \frac{3991}{24576}} =\frac{1215}{3991} \end{aligned}

Here’s the rest of the Bayes’ calculation:

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 0 \bold )
\displaystyle \begin{aligned}(21a) \ \ \ \ \ P(X=1 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{3}{4}}{\displaystyle \frac{10101}{24576}} =\frac{3072}{10101} \end{aligned}

\displaystyle \begin{aligned}(21b) \ \ \ \ \ P(X=2 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{9}{16}}{\displaystyle \frac{10101}{24576}} =\frac{2304}{10101} \end{aligned}

\displaystyle \begin{aligned}(21c) \ \ \ \ \ P(X=3 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{27}{64}}{\displaystyle \frac{10101}{24576}} =\frac{1728}{10101} \end{aligned}

\displaystyle \begin{aligned}(21d) \ \ \ \ \ P(X=4 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{81}{256}}{\displaystyle \frac{10101}{24576}} =\frac{1296}{10101} \end{aligned}

\displaystyle \begin{aligned}(21e) \ \ \ \ \ P(X=5 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{243}{1024}}{\displaystyle \frac{10101}{24576}} =\frac{972}{10101} \end{aligned}

\displaystyle \begin{aligned}(21f) \ \ \ \ \ P(X=6 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{729}{4096}}{\displaystyle \frac{10101}{24576}} =\frac{3729}{10101} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 1 \bold )
\displaystyle \begin{aligned}(22a) \ \ \ \ \ P(X=1 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{4}}{\displaystyle \frac{9094}{24576}} =\frac{1024}{9094} \end{aligned}

\displaystyle \begin{aligned}(22b) \ \ \ \ \ P(X=2 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{6}{16}}{\displaystyle \frac{9094}{24576}} =\frac{1536}{9094} \end{aligned}

\displaystyle \begin{aligned}(22c) \ \ \ \ \ P(X=3 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{27}{64}}{\displaystyle \frac{9094}{24576}} =\frac{1728}{9094} \end{aligned}

\displaystyle \begin{aligned}(22d) \ \ \ \ \ P(X=4 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{108}{256}}{\displaystyle \frac{9094}{24576}} =\frac{1728}{9094} \end{aligned}

\displaystyle \begin{aligned}(22e) \ \ \ \ \ P(X=5 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{405}{1024}}{\displaystyle \frac{9094}{24576}} =\frac{1620}{9094} \end{aligned}

\displaystyle \begin{aligned}(22f) \ \ \ \ \ P(X=6 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{1458}{4096}}{\displaystyle \frac{9094}{24576}} =\frac{1458}{9094} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 2 \bold ) done earlier

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 3 \bold )
\displaystyle \begin{aligned}(23a) \ \ \ \ \ P(X=3 \lvert Y=3)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{64}}{\displaystyle \frac{1156}{24576}} =\frac{64}{1156} \end{aligned}

\displaystyle \begin{aligned}(23b) \ \ \ \ \ P(X=4 \lvert Y=3)&=\frac{\displaystyle \frac{1}{6} \times \frac{12}{256}}{\displaystyle \frac{1156}{24576}} =\frac{192}{1156} \end{aligned}

\displaystyle \begin{aligned}(23c) \ \ \ \ \ P(X=5 \lvert Y=3)&=\frac{\displaystyle \frac{1}{6} \times \frac{90}{1024}}{\displaystyle \frac{1156}{24576}} =\frac{360}{1156} \end{aligned}

\displaystyle \begin{aligned}(23d) \ \ \ \ \ P(X=6 \lvert Y=3)&=\frac{\displaystyle \frac{1}{6} \times \frac{540}{4096}}{\displaystyle \frac{1156}{24576}} =\frac{540}{1156} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 4 \bold )
\displaystyle \begin{aligned}(24a) \ \ \ \ \ P(X=4 \lvert Y=4)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{256}}{\displaystyle \frac{211}{24576}} =\frac{16}{211} \end{aligned}

\displaystyle \begin{aligned}(24b) \ \ \ \ \ P(X=5 \lvert Y=4)&=\frac{\displaystyle \frac{1}{6} \times \frac{15}{1024}}{\displaystyle \frac{211}{24576}} =\frac{60}{211} \end{aligned}

\displaystyle \begin{aligned}(24c) \ \ \ \ \ P(X=6 \lvert Y=4)&=\frac{\displaystyle \frac{1}{6} \times \frac{135}{4096}}{\displaystyle \frac{211}{24576}} =\frac{135}{211} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 5 \bold )
\displaystyle \begin{aligned}(25a) \ \ \ \ \ P(X=5 \lvert Y=5)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{1024}}{\displaystyle \frac{22}{24576}} =\frac{4}{22} \end{aligned}

\displaystyle \begin{aligned}(25b) \ \ \ \ \ P(X=6 \lvert Y=5)&=\frac{\displaystyle \frac{1}{6} \times \frac{18}{1024}}{\displaystyle \frac{22}{24576}} =\frac{18}{22} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 6 \bold )
\displaystyle \begin{aligned}(26) \ \ \ \ \ P(X=6 \lvert Y=6)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{4096}}{\displaystyle \frac{1}{24576}} =1 \end{aligned}

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Probem 2
Let X be the value of one roll of a fair die. If the value of the die is x, we are given that Y \lvert X=x has a binomial distribution with n=x and p=\frac{1}{2} (we use the notation Y \lvert X=x \sim \text{binom}(x,\frac{1}{2})).

  1. Discuss how the joint probability function P[X=x,Y=y] is computed for x=1,2,3,4,5,6 and y=0,1, \cdots, x.
  2. Compute the conditional binomial distributions Y \lvert X=x where x=1,2,3,4,5,6.
  3. Compute the marginal probability function of Y and the mean and variance of Y.
  4. Compute P(X=x \lvert Y=y) for all applicable x and y.

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Answers to Probem 2

Problem 2.3

\displaystyle \begin{aligned} P(Y=y): \ \ \ \ &P(Y=0)=\frac{63}{384} \\&\text{ } \\&P(Y=1)=\frac{120}{384} \\&\text{ } \\&P(Y=2)=\frac{99}{384} \\&\text{ } \\&P(Y=3)=\frac{64}{384} \\&\text{ } \\&P(Y=4)=\frac{29}{384} \\&\text{ } \\&P(Y=5)=\frac{8}{384} \\&\text{ } \\&P(Y=6)=\frac{1}{384} \end{aligned}

\displaystyle E(Y)=\frac{7}{4}=1.75

\displaystyle Var(Y)=\frac{77}{48}

Problem 2.4

\displaystyle \begin{aligned} P(X=x \lvert Y=0): \ \ \ \ &P(X=1 \lvert Y=0)=\frac{32}{63} \\&\text{ } \\&P(X=2 \lvert Y=0)=\frac{16}{63} \\&\text{ } \\&P(X=3 \lvert Y=0)=\frac{8}{63} \\&\text{ } \\&P(X=4 \lvert Y=0)=\frac{4}{63} \\&\text{ } \\&P(X=5 \lvert Y=0)=\frac{2}{63} \\&\text{ } \\&P(X=6 \lvert Y=0)=\frac{1}{63} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=1): \ \ \ \ &P(X=1 \lvert Y=1)=\frac{32}{120} \\&\text{ } \\&P(X=2 \lvert Y=1)=\frac{32}{120} \\&\text{ } \\&P(X=3 \lvert Y=1)=\frac{24}{120} \\&\text{ } \\&P(X=4 \lvert Y=1)=\frac{16}{120} \\&\text{ } \\&P(X=5 \lvert Y=1)=\frac{10}{120} \\&\text{ } \\&P(X=6 \lvert Y=1)=\frac{6}{120} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=2): \ \ \ \ &P(X=2 \lvert Y=2)=\frac{16}{99} \\&\text{ } \\&P(X=3 \lvert Y=2)=\frac{24}{99} \\&\text{ } \\&P(X=4 \lvert Y=2)=\frac{24}{99} \\&\text{ } \\&P(X=5 \lvert Y=2)=\frac{20}{99} \\&\text{ } \\&P(X=6 \lvert Y=2)=\frac{15}{99} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=3): \ \ \ \ &P(X=3 \lvert Y=3)=\frac{8}{64} \\&\text{ } \\&P(X=4 \lvert Y=3)=\frac{16}{64} \\&\text{ } \\&P(X=5 \lvert Y=3)=\frac{20}{64} \\&\text{ } \\&P(X=6 \lvert Y=3)=\frac{20}{64} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=4): \ \ \ \ &P(X=4 \lvert Y=4)=\frac{4}{29} \\&\text{ } \\&P(X=5 \lvert Y=4)=\frac{10}{29} \\&\text{ } \\&P(X=6 \lvert Y=4)=\frac{15}{29} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=5): \ \ \ \ &P(X=5 \lvert Y=5)=\frac{2}{8} \\&\text{ } \\&P(X=6 \lvert Y=5)=\frac{6}{8} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=6): \ \ \ \ &P(X=6 \lvert Y=6)=1 \end{aligned}