{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# quiz8 - E 1 ∩ E 2 = P E 1 P E 2 = 1 12 and thus they are...

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

Name: Math 55 Quiz 8 SOLUTIONS July 31, 2009 GSI: Rob Bayer You have until 4:00 to complete this quiz. You must show your work. 1. (3 pts) Suppose you generate a bit string of length 10 uniformly at random. What is the probability that it either starts with a 0 or ends with a 110? There are 2 9 + 2 7 - 2 6 such strings, so the probability is 2 9 + 2 7 - 2 6 2 1 0 = 9 16 2. (3 pts) Consider the experiment in which we roll two fair dice. Determine whether the events “the sum of the dice is even” and “the second is a 4” are independent. Let E 1 be the event “the sum is even” and E 2 be the event “the second is a 4” For each of the 6 possible numbers for the first die, there are 3 that will make the sum even, so P ( E 1 ) = 18 36 = 1 2 Note P ( E 2 ) = 1 6 E 1 E 2 = { 24 , 44 , 64 } so P ( E 1 E 2 ) = 3 36 = 1 12
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: E 1 ∩ E 2 ) = P ( E 1 ) P ( E 2 ) = 1 12 and thus they are independent. 3. (4 pts) Suppose you ﬂip a bias-p coin four times and it comes up heads exactly twice. What’s the probability that the second ﬂip was a tails? Recall that a bias-p coin is one that comes up heads with probability p . Let E be the event “It comes up head exactly twice” and F be the event ”the second is a Tails.” So we want P ( F | E ) = P ( E ∩ F ) P ( E ) By Bernoulli, P ( E ) = ( 4 2 ) p 2 (1-p ) 2 . E ∩ F = { HTHT,HTTH,TTHH } . Each outcome here has probability p 2 (1-p ) 2 so P ( E ∩ F ) = 3 p 2 (1-p 2 ) and we have P ( F | E ) = 3 p 2 (1-p ) 2 6 p 2 (1-p ) 2 = 1 2...
View Full Document

{[ snackBarMessage ]}