CS206 -Discrete Mathematics II
Instructor: Chitoor V.Srinivasan.
PROBLEM SET 7 SOLUTIONS

Bayes' Theorem

(2).
Let   tex2html_wrap_inline110   denote ``someone tested is actually innocent,''   tex2html_wrap_inline112   denote ``someone tested is actually guilty,'' and let   tex2html_wrap_inline114   denote ``polygraph says he or she is innocent,'' let   tex2html_wrap_inline116   denote ``polygraph says he or she is guilty.'' Thus we have   tex2html_wrap_inline118 ,  and we get   tex2html_wrap_inline120 ,  and using Bayes' theorem, we get

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(3).
Let   tex2html_wrap_inline110   denote ``the car's oil pressure is too low,'' let   tex2html_wrap_inline112   denote ``the car's oil pressure is normal,'' and let   tex2html_wrap_inline114   denote ``the light flashes,'' let   tex2html_wrap_inline116   denote ``the light doesn't flash.'' Thus we have   tex2html_wrap_inline130 ,  and we get   tex2html_wrap_inline132 ,  and using Bayes' theorem, we get

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(5).
Let   tex2html_wrap_inline110   denote ``your next-door neighborhood is burglarized,'' let   tex2html_wrap_inline112   denote ``your next-door neighborhood isn't burglarized,'' and let   tex2html_wrap_inline114   denote ``the alarm sounds,'' let   tex2html_wrap_inline116   denote ``the alarm doesn't sound.'' Thus we have   tex2html_wrap_inline142 ,  and we can get   tex2html_wrap_inline144 ,  and using Bayes' theorem, we get

eqnarray21

(6).
Let   tex2html_wrap_inline110   denote ``Fracesca was enrolled in Prof.X's section,'' let   tex2html_wrap_inline112   denote ``Fracesca was enrolled in Prof.Y's section,'' and let   tex2html_wrap_inline114   denote ``she passed the course,'' let   tex2html_wrap_inline116   denote ``she failed in this course.'' Thus we have   tex2html_wrap_inline154 ,  and we can get   tex2html_wrap_inline156 ,  and using Bayes' theorem, we get

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(8).
According to Bayes' Theorem

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and that the fair coin tossed is a Bernoulli trial. Thus we have

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and we have   tex2html_wrap_inline158 . Hence we get
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(9).
Let   tex2html_wrap_inline110   denote ``the prisoner is related to a member of the governer's staff,'' let   tex2html_wrap_inline112   denote ``the prisoner isn't related to any member of the govner's staff,'' and let   tex2html_wrap_inline114   denote ``the prisoner is pardoned,'' let   tex2html_wrap_inline116   denote ``the prisoner isn't pardoned.'' Thus we have   tex2html_wrap_inline168 ,  and we can get   tex2html_wrap_inline170 ,  and using Bayes' theorem, we get

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(10).
Let   tex2html_wrap_inline110   denote ``Basil asked for the cherries flambe,'' let   tex2html_wrap_inline112   denote ``Basil asked for the chocolate mousse,'' let   tex2html_wrap_inline176   denote ``Basil asked for something else,'' and let   tex2html_wrap_inline114   denote ``Basil died,''   tex2html_wrap_inline116   denote ``Basil survived.'' Thus we have   tex2html_wrap_inline182 ,  and we get

eqnarray43

(12).
Let   tex2html_wrap_inline110   denote ``a child is actually abused,'' let   tex2html_wrap_inline112   denote ``a child isn't actually abused,'' and let   tex2html_wrap_inline114   denote ``the screening program diagnoses a child as abused,'' let   tex2html_wrap_inline116   denote ``the screening program diagnoses a child as nonabused.'' Thus we have,   tex2html_wrap_inline192 ,  and we get   tex2html_wrap_inline194 ,  and using Bayes' theorem, we get

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When   tex2html_wrap_inline110   is changed to  1/1000,  we will have   tex2html_wrap_inline200   and when   tex2html_wrap_inline110   is changed to  1/50  we will have   tex2html_wrap_inline206 .

Independence

(1).
Lets assume they have  N  black women, hence the probability of an employee is white is  (50+40)/(50+40+30+N),  and the probability of an employee is a man is  (50+30)/(50+40+30+N),  and the probability of an employee is a white man is  50/(50+40+30+N). If the company want to claim that the race and sex of the people on their staff are independent, they shall have

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solving for N yields  N=24. The claim that the race and sex of the people on their staff are independent is not equivalent to saying that they hire people independent of their race and sex, since the demographics of their staff may not reflect the demographics of their applicant pool.

(2).
Since   tex2html_wrap_inline218 ,  it illustrates that the event that Spike passes both of chemisty and mathematics isn't independent of the event that Spike passes chemistry and the event that Spike passes mathematics. Let  A  denote the event that Spike passes mathematics, and let  B  denote the event that Spike passes chemistry, hence the probability that Spike fails both is

eqnarray58

(5).
No, the chance of someone else intends to blow up isn't lessened by the fact that you bring a bomb on board the plane, since they are independent events.

(6).
Since the events   tex2html_wrap_inline224   are mutually independent, we have   tex2html_wrap_inline226 ,  and   tex2html_wrap_inline228 . Thus, we get

eqnarray63

Hence we have   tex2html_wrap_inline230 . Therefore   tex2html_wrap_inline232   and  C  are independent.

(8).
Each die has outcomes   tex2html_wrap_inline236 . The sample space is a set of ordered pairs   tex2html_wrap_inline238 .  Event tex2html_wrap_inline240  , event   tex2html_wrap_inline242 ,  and event   tex2html_wrap_inline244 P(A) = P(B) = 18/36 = 1/2.  Now   tex2html_wrap_inline248 Thus, A and B are pairwise independent. tex2html_wrap_inline250 .  Now  tex2html_wrap_inline252   and   tex2html_wrap_inline254   Thus the pairs (A,C) and (B,C) are also pairwise independent. But since   tex2html_wrap_inline256   A, B, and C are not mutually independent.

(9).
Lets assume   tex2html_wrap_inline258 . Assume, events   tex2html_wrap_inline260   for   tex2html_wrap_inline262   are pairwise independent and   tex2html_wrap_inline264 . Let   tex2html_wrap_inline266   be the probability of   tex2html_wrap_inline268 . Clearly,   tex2html_wrap_inline270 .

Since   tex2html_wrap_inline264tex2html_wrap_inline274   for any  i,    tex2html_wrap_inline278 . Thus each   tex2html_wrap_inline280   is either a singleton set or a doubleton set. Clearly,   tex2html_wrap_inline282   for   tex2html_wrap_inline284 ,  since if   tex2html_wrap_inline286 ,  then we will have,   tex2html_wrap_inline288 ,  and in this case   tex2html_wrap_inline280   and   tex2html_wrap_inline292   will not be independent of each other, contradicting our assumption.

Now, for any pair   tex2html_wrap_inline280   and   tex2html_wrap_inline292   there are only two possibilities: Either   tex2html_wrap_inline298 ,  or   tex2html_wrap_inline300 . If   tex2html_wrap_inline298 ,  then we will have,   tex2html_wrap_inline304 ,  and this will make   tex2html_wrap_inline280   and   tex2html_wrap_inline292   not independent of each other, again contradicting our assumption.

If   tex2html_wrap_inline300 ,  since   tex2html_wrap_inline312   and   tex2html_wrap_inline314   are both true, both   tex2html_wrap_inline280   and   tex2html_wrap_inline292   should be doubleton sets such that   tex2html_wrap_inline320   for some   tex2html_wrap_inline322tex2html_wrap_inline324tex2html_wrap_inline326 and   tex2html_wrap_inline328   for distinct   tex2html_wrap_inline330   and  k. In this case, the following will hold true:

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and, since   tex2html_wrap_inline280   and   tex2html_wrap_inline292   are independent,

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Thus we get   tex2html_wrap_inline338   and hence   tex2html_wrap_inline340 ,  which is impossible since   tex2html_wrap_inline342   is true. Hence, it is impossible for   tex2html_wrap_inline344   and   tex2html_wrap_inline346   to be pairwise independent.

(12).
Since   tex2html_wrap_inline348   are independent events,

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(13).
Lets give an example such that   tex2html_wrap_inline350   isn't equal to   tex2html_wrap_inline352  .

Let   tex2html_wrap_inline224   be the events in Problem 8 above. We know that  A  and  B  are independent, and  P(C) > 0  and we have   tex2html_wrap_inline362 . But,   tex2html_wrap_inline364  . Thus we prove that   tex2html_wrap_inline350   is not necessarily equal to   tex2html_wrap_inline352 .