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Solutions to the HW4 problems (from lecture 5)

Solutions to the HW4 problems (from lecture 5).

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Solutions to the HW4 problems (from lecture 5)

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  1. Solutions to the HW4 problems(from lecture 5)

  2. Problem 1: Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 2 white balls. A ball is chosen at random from the first box and placed in the second box without observing its color. Then the ball is drawn from the second box. Find the probability that it is white. Solution: W1=“a white ball was chosen from the Box 1”; P(W1) = 5/8;R1 = “a red ball was chosen from the Box 1” ; P(R1) = 3/8;W2 = “a white ball was chosen from the Box 2” W2 can be presented as a union of two mutually exclusive events P(W2) = P(W1)*P(W2|W1) + P(R1)*P(W2|R1)= (5/8)* (3/7) + (3/8)*(2/7)= 21/56 = 3/8.

  3. Problem 2: A friend rolls two dice and tells you that there is at least one 6. What is the probability that the sum is at least 9?A= “ at least one 6”= orange + yellow; P(A) = 11/36. B= “ the sum is ay least 9”= red + green. AB= red; P(AB) =7/36.P(B|A) = P(AB)/P(A)= 7/11.

  4. Some difficulties were caused by the “cards” problem. Problem 3We draw 4 cards out of deck of 52. Find the probability that all four values are different given that (a) all four cards belong to different suits (b) We drew two spades and two hearts. case (a). Events: B=“All four suits are different”; A=“All cards are different”.P(B) = 134/C52,4=0.1055; P(AB)= 13*12*11*10/ C52,4=0.0634 P(A|B) = P(AB)/P(B)=0.601 Case (b) B= “Two spades and two hearts”. P(B) =C13,2C13.2/C52,4. P(AB) =C13,2C11,2/C52,4; P(A|B) = P(AB)/P(B)=0.705

  5. Problem 4 An urn contains 8 red, 7 blues and 5 green balls. You draw out two balls and they are different color. What is the probability that the two balls were red and blue. B=“The balls have different color”; A =“ One is red and the other is blue” P(B) = 1- p(Bc) = 1- [P(both red) + P(both blue) + P(both green)]=1- (8/20)(7/19)-(7/20)(6/19)-(5/20)(4/19) = 131/190. P(A)=(remember that the order can be different !) = (8/20)(7/19)+(7/20)(8/19) = 28/95. Notice now that AB, meaning that AB=A.As a result, P(A|B)=P(A)/P(B)= 56/131= 0.428

  6. Problem 5: North flips 3 coins and South flips 2. North wins if the number of tails he gets is more than the number South gets. What is the probability North will win? Let A=" North wins ", Sn=“South gets n tails", Cm = “North gets m tails".P(S0)= ¼. P(S1)=1/2, P(S2)=1/4 P(N0) = 1/8 = P(N3); P(N1))= 3/8 =P(N2). P(m>n)= P(S0) *(P(N1)+P(N2)+P(N3) ) + P(S1) * (P(N2)+P(N3) ) + P(S2)*P(N3) = 1/4*7/8+ 1/2)*1/2+1/4*1/8= ½.

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