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 2
nd
and 3
rd
pick 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, 15
2
).

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