X/Y | 20 | 30 | 40 | 50 |

4 | 0.01 | 0.03 | 0.05 | 0.02 |

5 | 0.03 | 0.1 | 0.08 | 0.04 |

6 | 0.02 | 0.08 | 0.12 | 0.11 |

7 | 0.02 | 0.04 | 0.07 | 0.18 |

- Find the marginal probability distribution for X and Y
- Determine the conditional probability distribution of Y gives X = 5
- Are x and Y independent

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X/Y |
20 |
30 |
40 |
50 |
P(X=x) |

4 |
0.01 | 0.03 | 0.05 | 0.02 | 0.11 |

5 |
0.03 | 0.1 | 0.08 | 0.04 | 0.25 |

6 |
0.02 | 0.08 | 0.12 | 0.11 | 0.33 |

7 |
0.02 | 0.04 | 0.07 | 0.18 | 0.31 |

P(Y=y) |
0.08 | 0.25 | 0.32 | 0.35 | 1 |

Now,

Marginal probability of X = p(X = x_{i}) = \(\sum_{i=4}^n p(x_i y_i)\)

∴P(X = 4) = P(X=4, Y=20) + P(X = 4, Y = 30) + P(X = 4, Y = 40) + P(X = 4, Y = 50)

= 0.11

∴P(X = 5) = P(X=5, Y=20) + P(X = 5, Y = 30) + P(X = 5, Y = 40) + P(X = 5, Y = 50)

= 0.25

∴P(X = 6) = P(X=6, Y=20) + P(X = 6, Y = 30) + P(X = 6, Y = 40) + P(X = 6, Y = 50)

= 0.33

∴P(X = 7) = P(X=7, Y=20) + P(X = 7, Y = 30) + P(X = 7, Y = 40) + P(X = 7, Y = 50)

= 0.31

Marginal probability of Y = p(Y = y_{i}) = \(\sum_{i=4}^n p(x_i y_i)\)

∴P(Y = 20) = P(X=4, Y=20) + P(X = 5, Y = 20) + P(X = 6, Y = 20) + P(X = 7, Y = 20)

= 0.08

Similarly,

P(Y = 30) = 0.25

P(Y = 40) = 0.32

P(Y = 50) = 0.35

**Answer b):**

The conditional probablity function of Y given the value of X = 5 is designed as follow:

P(X=5 / Y = 20) = \(\frac{P(X=5 \enspace / \enspace Y = 20)}{P(Y = 20)}\)

= \(\frac{0.03}{0.08} = \frac{3}{8}\)

P(X=5 / Y = 30) = \(\frac{P(X=5 \enspace / \enspace Y = 30)}{P(Y = 30)}\)

= \(\frac{0.1}{0.25} = \frac{2}{5}\)

P(X=5 / Y = 40) = \(\frac{P(X=5 \enspace / \enspace Y = 40)}{P(Y = 40)}\)

= \(\frac{0.08}{0.32} = \frac{1}{4}\)

P(X=5 / Y = 50) = \(\frac{P(X=5 \enspace / \enspace Y = 50)}{P(Y = 50)}\)

= \(\frac{0.04}{0.35} = \frac{4}{35}\)\

**Answer c):**

Y is independent variable.

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