{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

324.midterm01.2005.solution

# 324.midterm01.2005.solution - 2005 STAT 324 midterm 1...

This preview shows pages 1–2. Sign up to view the full content.

2005 STAT 324 midterm 1 solution Moo K. Chung [email protected] Grade distribution 29 . 65 ± 6 . 18. 15 20 25 30 35 40 0 1 2 3 4 5 6 1. (a) Suppose that A and B are independent events such that the probability that neither occurs is 1 2 and the probability of B occurring is 1 3 . Determine the probability of A occurring. Solution. P (( A B ) c ) = 1 - P ( A B ) = 1 / 2. So P ( A B ) = 1 / 2. Since A and B are independent P ( A B ) = P ( A ) P ( B ) = P ( A ) / 3. By substituting terms into P ( A B ) = P ( A ) + P ( B ) - P ( A B ), we have 1 / 2 = P ( A ) + 1 / 3 - P ( A ) / 3. Hence P ( A ) = 1 / 4. (b) If A and B are independent events, prove that A c and B are also independent. Solution. Since A and B are independent, P ( A B ) = P ( A ) P ( B ). We need show P ( A c B ) = P ( A c ) P ( B ). P ( B ) = P ( A B )+ P ( A c B ) = P ( A ) P ( B )+ P ( A c B ). Hence P ( A c B ) = P ( B ) - P ( A ) P ( B ) = (1 - P ( A )) P ( B ) = P ( A c ) P ( B ). 2. A lie detector will show a positive reading (indicates a lie) 10% of the time when a person is telling the truth and 90% of the time when the person is lying. Suppose two people are suspects in a crime and only one is guilty. The guilty one will be

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

324.midterm01.2005.solution - 2005 STAT 324 midterm 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online