Download 26 Exercises 8, 20, 24, 26, 28 8. A company has only one position

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Homework Assignment 2:
Pages 23 - 26 Exercises 8, 20, 24, 26, 28
8. A company has only one position with three highly qualified applicants: John,
Barbara, and Marty. However, because the company has only a few women
employees, Barbara's chance to be hired is 20% higher than John's and 20% higher
than Marty's. Find the probability that Barbara will be hired.
Solution: Let P_b, P_j, P_m be the respective probabilities that Barbara, John, and
Marty will be hired. Since the events of hiring hiring Barbara, John, or Marty are
pairwise disjoint and jointly exhaustive, we must have that
P_b + P_j + P_m = 1.
We also know that P_j = P_b - 0.2 and P_m = P_b - 0.2. Substituting these into the
previous equation gives 3 P_b - 0.4 =1 and therefore
P_b = 1.4/3 = 14/30
which gives the approximate value of 0.467 for Barbara's probability of being hired.
20. The coefficients of the quadratic equation x^2 + b x + c = 0 are determined by
tossing a fair die twice (the first outcome is b, the second one is c). Find the
probability that the equation has real roots.
Solution: The sample space consists of 36 equally likely ordered pairs ( b, c ), where
b and c take on any of the values 1, 2, 3, 4, 5, 6. The quadratic equation has real
roots if and only if b^2 - 4 c is greater than or equal to 0. The condition is satisfied by
19 of the 36 ordered pairs, thus the probability that the quadratic equation has real
root is 19/36, which is approximately equal to 0.528.
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24. From an ordinary deck of 52 cards, we draw cards at random and without
replacement until only cards of one suit are left. Find the probability that the cards left
are all spades.
Solution. Let P_d, P_c, P_h, P_s be the respective probabilities that the only cards left
are diamonds, clubs, hearts, or spades. Since the four events of having only
diamonds, only clubs, only hearts, or only spades are pairwise mutually exclusive
and jointly exhaustive, we must have that
P_d + P_c + P_h + P_s = 1.
Moreover, since each suit has the same number of cards each of the four events
mentioned above is equally likely to occur, and therefore we must also have that
P_d = P_c = P_h = P_s,
which implies that 4 P_s = 1. Thus P_s = 1/4 = 0.25.
26. For a Democratic candidate to win an election, she must win districts I, II, and III.
Polls have shown that the probability of winning I and III is 0.55, losing II but not I is
0.34, and losing II and III but not I 0.15. Find the probability that this candidate will
win all three districts. (Draw a Venn diagram.)
Figure 1.
Winning I and III
Prob = 0.55
Figure 2.
Losing II but not I
Prob = 0.34
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Figure 3.
Losing II and III but not I
Prob = 0.15
Figure 4.
Figure 2 - Figure 3
Prob = 0.34 - 0.15 = 0.19
In these Venn diagrams, the red circle
represents winning district I, the blue circle
represents winning district II, and the green
circle represents winning district III.
The diagrams show that the probability of
winning all three districts is 0.36.
Figure 5.
Figure 1 - Figure 4
Prob = 0.55 - 0.19 = 0.36
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28. Let A_1, A_2, A_3, ... be a sequence of events of a sample space. Prove that