Hw2solalt

Hw2solalt - EE 178 Probabilistic Systems Analysis Homework...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
EE 178 Handout #5 Probabilistic Systems Analysis January 29, 2008 Homework Solutions #2 1. (10 points) a. Let R be the event corresponding to Riddley running and let T be the event corre- sponding to Riddley catching the train. The tree is shown in Figure 1. b. Using the law of total probability, P( T ) = P( T | R )P( R ) + P( T | R c )P( R c ) = 2 5 × 2 5 + 1 2 × 3 5 = 0 . 46 . c. Using Bayes’ rule, P( R | T ) = P( T | R )P( R ) P( T ) = 0 . 3478 . d. The probability of catching the train on any given day is 0.46. Therefore, the proba- bility of catching the train on any 2 days out of 5 is p 5 2 P 0 . 46 2 (1 0 . 46) 3 = 0 . 3332 . 2. (10 points) a. Let D be the event that the person has the disease and let T be the event that the test is positive. Then, P( D ) = 1 / 20, P( T c | D ) = 1 / 50, and P( T | D c ) = 1 / 10. Then, using the Bayes rule, P( D | T ) = P( T | D )P( D ) P( T | D )P( D ) + P( T | D c )P( D c ) = (49 / 50)(1 / 20) (49 / 50)(1 / 20) + (1 / 10)(19 / 20) = 0 . 3403 . b. Let F be the event that the father has the disease and let S be the event that the son has the disease. Then, P( F ) = 1 / 20, P( S | F ) = 4 / 5, and P( S | F c ) = 1 / 95. Then, P( F | S ) = P( S | F )P( F ) P( S | F )P( F ) + P( S | F c )P( F c ) = (4 / 5)(1 / 20) (4 / 5)(1 / 20) + (1 / 95)(19 / 20) = . 80 . 3. (10 points) By the total law of probability, P( A ) = P( A | B )P( B ) + P( A | B c )P( B c ) .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
R R c 2 / 5 3 / 5 T T c T T c 2 / 5 3 / 5 1 / 2 1 / 2 Figure 1: Probability tree for Riddley’s morning. Page 2 of 8 EE 178, Winter 2008
Background image of page 2
Therefore P( A | B c )P( B c ) = P( A ) P( A | B )P( B ) . We can bound P( A | B c ) as follows: P( A | B c ) = P( A ) P( A | B )P( B ) P( B c ) P( A ) P( A )P( B ) P( B c ) since P( A | B ) P( A ) = P( A )(1 P( B )) 1 P( B ) = P( A ) . 4. (15 points) a. Let A i be the event that the input is i and B i be the event that the output is i . Using the law of total probability and conditional probability we get P( B 0 ) = P( B 0 | A 0 )P( A 0 ) + P( B 0 | A 2 )P( A 2 ) = 1 2 (1 ǫ ) + 1 4 ǫ = 1 2 1 4 ǫ, P( B 1
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

Hw2solalt - EE 178 Probabilistic Systems Analysis Homework...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online