HW3 - Section 3.1 - 3.2 Random variables 3.3 Solution: The...

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Section 3.1 - 3.2 Random variables 3.3 Solution: The sample space S for the three tosses of the coin is: S = { HHH,HHT,HTH,HTT,THH,THT,TTH,TTT } Let W be a random variable giving the number of heads minus the number of tails in three tosses of a coin, we assign a value of ω of W to each sample point in the following way: Sample points ω HHH 3 HHT 1 HTH 1 HTT -1 THH 1 THT -1 TTH -1 TTT -3 3.8 Solution: Since the coin is biased, the probability of getting a head for one toss is 2/3 and the prob. of getting a tail is 1/3. From 3.3, we can get the probability distribution function (p.d.f.): P(x=-1)=P(HTT)+P(THT)+P(TTH)=3 · ( 2 3 · 1 3 · 1 3 ) = 2 9 P(x=1)=P(HHT)+P(HTH)+P(THH)=3 · ( 2 3 · 2 3 · 1 3 ) = 4 9 P(x=-3)=P(TTT)= 1 3 · 1 3 · 1 3 = 1 27 P(x=3)=P(HHH)= 2 3 · 2 3 · 2 3 = 8 27 3.11 Solution: If X is the number of defective sets purchased by the hotel, then X can be 0, 1, 2. The p.d.f. of X is: P(X=0)= ( 5 3 ) ( 7 3 ) = 2 7 P(X=1)= ( 2 1 )( 5 2 ) ( 7 3 ) = 4 7 P(X=2)= ( 5 1 ) ( 7 3 ) = 1 7 1
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3.12 Solution: (a) P(T=5)=F(5)-F(4)=3/4 - 1/2=1/4 (b) P ( T 3) = 1 - P ( T 3) = 1 - F (3) = 1 - 1 / 2 = 1 / 2 (c) P (1 . 4 < T < 6) = P ( T 6) - P ( T 4) - P ( T = 6) = F (6) - F (1 . 4) - ( F (6) - F (5))=3/4 - 1/4 - (3/4 - 3/4)=1/2
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HW3 - Section 3.1 - 3.2 Random variables 3.3 Solution: The...

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