Refer to Exercise 4.

a. Find the conditional probability mass function \(p_{Y \mid X}(y \mid 1)\).

b. Find the conditional probability mass function \(p_{X \mid Y}(x \mid 2)\).

c. Find the conditional expectation \(E(Y\mid X=1)\).

d. Find the conditional expectation \(E(X\mid Y=2)\).

Equation Transcription:

Text Transcription:

p_{Y|X}(y|1)

p_{X|Y}(x|2)

E(Y|X=1)

E(X|Y=2)

Step 1 of 5:

Here,it is given that X be the number with too little clearance and Y be the number with too much clearance.

Also the joint probability mass function of X and Y is given as

Y | ||||

X |
0 |
1 |
2 |
3 |

0 |
0.15 |
0.12 |
0.11 |
0.10 |

1 |
0.09 |
0.07 |
0.05 |
0.04 |

2 |
0.06 |
0.05 |
0.04 |
0.02 |

3 |
0.04 |
0.03 |
0.02 |
0.01 |

Using these,we have to find the required probabilities and expectations.