This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 3C Homework 3 Solutions Ilhwan Jo and Akemi Kashiwada [email protected], [email protected] Assignment: Section 12.3 Problems 2, 7, 8, 9, 11, 13, 15, 18, 21, 22, 29, 31, 32 2. You draw three cards from a standard deck of 52 cards. Find the probability that the third card is a club given that the first two cards were clubs. Solution Let A be the event that the third card is a club and let B the event that the first two cards were clubs. Then the probability that all three cards are clubs is P ( A ∩ B ) = ( 13 3 ) ( 52 3 ) , and the probability of the first two cards being clubs is P ( B ) = ( 13 2 ) ( 52 2 ) . So the conditional probability that the third card is a club given that the first two were clubs is P ( A  B ) = P ( A ∩ B ) P ( B ) = ( 13 3 ) ( 52 3 ) ( 13 2 ) ( 52 2 ) = 11 50 . Another way of doing this problem is noticing that the probability of the first card being a club is 13 52 and the probability of the second card being a club is 12 51 . As 2 clubs have already been removed from the deck, there are only 11 clubs left out of the 50 remaining cards. So the probability that the third card is a club given that the two we already removed are clubs is 11 50 . 7. You roll two fair dice. Find the probability that the first die is a 4 given that the sum is 7. Solution Let A denote the event that the first die shows 4 and B denote the event that the sum of the dice is 7. Notice that for any number the first die shows, there is only one number the second die can show to make the sum 7 (e.g. if the first die shows 5 then the second die must show 2 to make the sum 7). So there are a total of 6 ways to make the sum of the two dice equal 7 so P ( B ) = 6 36 . Now, if the first die shows 4 there is only one way to make the sum of both dice equal 7 which means P ( A ∩ B ) = 1 36 . Therefore, the probability that the first die shows 4 given that the sum is 7 is P ( A  B ) = P ( A ∩ B ) P ( B ) = 1 / 36 6 / 36 = 1 6 . 1 8. You roll two fair dice. Find the probability that the first die is a 5 given that the minimum of the two numbers is 3. Solution Let A denote the event that the first die shows 5 and B denote the event that the minimum of both dice is 3. If the minimum of both dice is 3, then at least one of the two dice must be 3 and both must be greater than or equal to 3. So P ( B ) = 7 36 . The probability that the first die shows 5 and the second shows 3 is P ( A ∩ B ) = 1 36 ....
View
Full Document
 Spring '07
 SCHONMANN
 Math, Conditional Probability, Probability, Probability theory, Playing card, Dice

Click to edit the document details