Review Problems Solutions

# Review Problems Solutions

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: heavy coﬀee drinker” and C be the event “having cancer of the pancreas.” Since we are testing for independence, we must compare P (H ∩ C ) with P (H )P (C ). If they are equal then the events are independent. If they are not equal then the events are not independent. 121 P (H ∩ C ) = 10,000 = .0121 P (H ) = 4,000 10,000 and P (C ) = 151 10,000 , so P (H )P (C ) = .00604. This means that P(H ∩ C) ￿= P(H)P(C) and so H and C are NOT INDEPENDENT. Problem 9 Copykwik has four photocopy machines A, B, C, and D. The probability that a given machine will break down on a particular day is given below. Assuming independence, what is the probability on a particular day that the following will occur? P (A) = .002 P (B ) = .003 P (C ) = .001 P (D) = .002 (A) None of the machines will break down. Solution: This means A does not break down AND B does not break down AND C does not break down AND D does not break down. So we want P (Ac ∩ B c ∩ C c ∩ Dc ). Since the events A, B , C , and D are independent we have P ( Ac ∩ B c ∩ C c ∩ D c ) = P ( Ac ) P ( B c ) P ( C c ) P ( D c ) = (1 − P (A))(1 − P (B ))(1 − P (C ))(1 − P (D)) = ≈ (1 − .002)(1 − .003)(1 − .001)(1 − .002) .9920 (B) All of the machines break down. Solution: P (A ∩ B ∩ C ∩ D) = P (A)P (B )P (C )P (D) = (.002)(.003)(.001)(.002) = 1.2 × 10−11 (C) Exactly one machine breaks down. Solution: We want to ﬁnd P ( A ∩ B c ∩ C c ∩ D c ) + P ( Ac ∩ B ∩ C c ∩ D c ) + P ( Ac ∩ B c ∩ C ∩ D c ) + P ( Ac ∩ B c ∩ C c ∩ D ) . P (A ∩ B c ∩ C c ∩ Dc ) = (.002)(1 − .003)(1 − .001)(1 − .002) = .001988021988 P (Ac ∩ B ∩ C c ∩ Dc ) = (1 − .002)(003)(1 − .001)(1 − .002) = .002985023988 P (Ac ∩ B c ∩ C ∩ Dc ) = (1 − .002)(1 − .003)(.001)(1 − .002) = .000993015988 P (Ac ∩ B c ∩ C c ∩ D) = (1 − .002)(1 − .003)(1 − .001)(.002) = .001988021988 So P ( A ∩ B c ∩ C c ∩ D c ) + P ( Ac ∩ B ∩ C c ∩ D c ) + P ( Ac ∩ B c ∩ C ∩ D c ) + P ( Ac ∩ B c ∩ C c ∩ D ) = .001988021988 + .002985023988 + .000993015988 + .001988021988 = .007954083952 ≈ .007954 (D) Machine...
View Full Document

## This document was uploaded on 03/17/2014 for the course MATH 1331 at Texas Tech.

Ask a homework question - tutors are online