hw4 - Solutions to Homework 4 - 3C 12.3.10 The question is...

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Unformatted text preview: Solutions to Homework 4 - 3C 12.3.10 The question is asking us to find P ( A | B ) where we let A = the event that at least two heads occurred, B = the event that the second toss was a head. We can find this using the formula P ( A | B ) = P ( A ∩ B ) /P ( B ). The sample space is Ω = { HHH,HHT,HTH,HTT,THH,THT,TTH,TTT } , and thus we can see that the sets we need are A ∩ B = { HHH,HHT,THH } and B = { HHH,HHT,THH,THT } . From these we can deduce that P ( A ∩ B ) = | A ∩ B | | Ω | = 3 8 , P ( B ) = | B | | Ω | = 1 2 , and hence P ( A | B ) = 3 / 4. 2.3.14 Let the A be the event that that the ball we pick is blue, and let B i be the event that we pick the ball out of bag i . e.g. B 1 is the event that we pick the ball out of bag 1. To find P ( A ) we will use the Law of Total Probability P ( A ) = 6 X i =1 P ( A | B i ) P ( B i ) . Therefore we need to find P ( A | B i ) and P ( B i ). Here’s another way of thinking of the B i : its the event that the value of the die-roll is i . Therefore we can deduce that P ( B i ) = 1 6 since the probability that we roll each value on a die is equal, and there are 6 sides. P ( A | B i ) represents the probability that we pick a blue ball out of bag i . Bag i contains i blue ball and 2 green balls (so i + 2 in total). Therefore we can deduce that P ( A | B i ) = i i + 2 . Putting this all together, we get P ( A ) = 1 6 1 3 + 1 2 + 3 5 + 2 3 + 5 7 + 3 4 = 0 . 59 12.3.23 1 Since no replacement is allowed after you take the first ball, we can intuitively guess that the two events are not independent. To show that this we need toguess that the two events are not independent....
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This note was uploaded on 04/03/2008 for the course MATH 3C taught by Professor Schonmann during the Spring '07 term at UCLA.

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hw4 - Solutions to Homework 4 - 3C 12.3.10 The question is...

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