CS206 -Discrete Mathematics II
Instructor: Chitoor V.Srinivasan
PROBLEM SET 5 SOLUTIONS

  1. After the first card, a queen, was drawn out from the packet of  52  cards, the sample space is decreased to  51 and there will still be  4  kings left, hence the probability the second card drawn is king, is   tex2html_wrap_inline131 .

  2. The event, the two numbers total 7, contains the following different sample outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). The size of the sample space is number of possible rolls minus the number of possible doubles   tex2html_wrap_inline133 . Thus, the probability is tex2html_wrap_inline135 .

  3. The probability of choosing each group is   tex2html_wrap_inline137 , and the probability of a man would be picked out from the first group, after the first group was chosen, is   tex2html_wrap_inline139 , and the probability of a man would be picked out from the second group, after the second group was chosen, is   tex2html_wrap_inline141 . Hence the probability a man would be picked out is   tex2html_wrap_inline143 .

  4. The probability of a person is a male or female is   tex2html_wrap_inline137 , and the chance of a male contracting measles is   tex2html_wrap_inline147 , and the chance of a female contracting measles is   tex2html_wrap_inline149 , hence the probability of a person contracting measles is   tex2html_wrap_inline151 .

  5. We draw cards from the pack of  52  cards, without replacement.
    (a).
    We have  13  clubs in the total  52  cards, hence the probability of the first card drawn is a club is   tex2html_wrap_inline159 .

    (b).
    Since the first card drawn is club, it decreases the whole sample space to  51  and the probability of the second card drawn is a club is   tex2html_wrap_inline163 . Hence the probability of both cards drawn are clubs is   tex2html_wrap_inline165 .

    (c).
    The probability of the second card drawn is a club is the sum of the probability of the both cards drawn are club, and the probability of the first card drawn is not a club and the second card drawn is a club, which is   tex2html_wrap_inline167 .

  6. If we have tested n parts, then the probability of finding the faulty part is n/100. We want n/100 to be .80. Thus, n=80 parts have to be tested.

  7. Let   tex2html_wrap_inline169   denote  `the first urn is chosen' , and   tex2html_wrap_inline171   denote  `the second urn is chosen' , and   tex2html_wrap_inline173   denote  `the third urn is chosen' , and  B  denote  `A ball drawn is red'. According to BAYES'THEOREM, we have

    displaymath121

    And we know   tex2html_wrap_inline177 , and   tex2html_wrap_inline179 , and   tex2html_wrap_inline181 , and   tex2html_wrap_inline183 , thus we get

    displaymath122

  8. The chance that the first component would fail in a week is  0.7, and the chance that the second component would fail in a week is  0.2, hence the chance that the first component would last in a week is  0.3, and the chance that the second component would last in a week is  0.8. Thus the chance that the machine lasts a week is   tex2html_wrap_inline193 .

  9. There is a   tex2html_wrap_inline195   chance of having a storm on any day, and the chance that no storm on any day is   tex2html_wrap_inline197 . Hence the chance that there is no storm on three consecutive days is   tex2html_wrap_inline199 .

  10. The chance he interviews at least once in a year is 1 minus the chance he gets no interviews in a year. Hence, the probability is tex2html_wrap_inline201