# 2.16 - 3 P(E 1 E 2 P(E 2 E 3 P(E 1 E 3 P(E 1 E 2 E 3 = 0.3...

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

2.16 a. E 1 = Monday is a rainy day E 2 = Tuesday is a rainy day E 3 = Wednesday is a rainy day Given P(E 1 ) = P(E 2 ) = P(E 3 ) = 0.3 P(E 2 E 1 ) = P(E 3 E 2 ) = 0.5 P(E 3 E 1 E 2 ) = 0.2 b. P(E 1 E 2 ) = P(E 2 E 1 ) P(E 1 ) = 0.5x0.3 = 0.15 c . P(E 1 E 2 3 E ) = P( 3 E E 1 E 2 ) P(E 2 E 1 ) P(E 1 ) = (1-0.2)(0.5)(0.3) = 0.12 d. P(at least one rainy day) = P(E 1 E 2 E 3 ) = P(E 1 ) + P(E 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3 ) - P(E 1 E 2 ) - P(E 2 E 3 ) - P(E 1 E 3 ) + P(E 1 E 2 E 3 ) = 0.3 + 0.3 + 0.3 - 0.3x0.5 - 0.3x0.5 – 0.1125 + 0.3x0.5x0.2 =0.52 Where P(E 1 E 3 )=P(E 3 |E 1 )P(E 1 )=0.375x0.3=0.1125 P(E 3 |E 1 )= P(E 3 |E 2 E 1 )P(E 2 |E 1 )+ P(E 3 | 2 E E 1 )P( 2 E |E 1 ) =0.15x0.15+0.3x0.5 =0.375...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online