ISMT111 F05 Midterm Exam Suggested solution Question 1 (a)One gets the right sequence by either the combination (B,R,R) or (R,R,R). The probability that a red ball is chosen on the 2ndand 3rdpick is then (3/8)(5/7)(4/6)+(5/8)(4/7)(3/6)=5/14 (b)Mean is 100(5/14) and variance is 100(5/14)(9/14) (c)Use P(Y100<=20.5) then apply the normal approximation. (d)This is binomial with n=10, use the binomial exact formula. (e)Let AR be the event that an extra red ball is added to a box. Then the probability of adding a red ball to a single box is P(AR)=P(AR|R)P(R)+P(AR|B)P(B)=1/2(5/8); This is a binomial problem and hence the expectation is 100(1/2) 5/8. Question 2 a)P(head)=1 x (1/2)+(1/2)(1/2)=3/4 b)P(coin 1 and head)=P(head|coin1)P(coin1)=(1/2)(1/2)=1/4 c)P(coin1|head)=[P(coin1, head)]/P(head)=1/3 P(coin2|head) )=[P(coin2, head)]/P(head)=2/3 d)P(head down|head up)=[P(head down and head up)]/[P(head)] = P(coin2)/P(head) =(1/2)/(3/4)=2/3 Or P(head down|head up)=P(coin2 |head)=2/3 Question 3: Let X be the number of minutes need to complete the exam, X~N(100, 152).
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