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

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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 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

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

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