1 - 1) A Question 1 p=0.8, prize=100 $ B Question 2 p=0.5,...

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A Question 1 p=0.8, prize=100 $ B Question 2 p=0.5, prize= 200 $ E[A] = 0*0.2 + 100*(0.8*0.5) + 300*(0.8*0.5) = 0 + 40 + 120 = 160 E[B]= 0*0.5 + 200*(0.5*0.2) + 300*(0.5*0.8) = 0 + 20 + 120 =140 As we see the expected value of A is greater than B. So first question should be chosen. 2) E: event that the man discovers the matchbox in his right pocket is empty and there are k matches in the matchbox in his left pocket. If the (N+1)th choice of the matchbox in his right pocket is made at the (N+1+N-k) trial, this event will occur. E is a random variable with the negative binomial distribution p=1/2 , r= N+1, n=2N-k+1 = - ( ) - + PE 2N kN 12 2N k 1 Also if we think the opposite of this event, which we can say that being empty of the matchbox in his left pocket, the probability will be equally likely as; = - ( ) - PE 2N kN 12 2N k 3) i (number of the winnings) p (the probability that A team wins) E[X] (expected number of the games) a) i=2 There can be totally 2 games or 3 games in this case. E[X] = (2*1! )*(p
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1 - 1) A Question 1 p=0.8, prize=100 $ B Question 2 p=0.5,...

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