CS206 -Discrete Mathematics II Instructor: Chitoor V.Srinivasan PROBLEM SET 5 SOLUTIONS
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 .
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 . Thus, the probability is
.
The probability of choosing each group is
, and the probability of a man would be picked out from
the first group, after the first group was chosen, is ,
and the probability of a man would be picked out from the second
group, after the second group was chosen, is . Hence
the probability a man would be picked out is
.
The probability of a person is a male or female is
, and the chance of a male contracting measles is
, and the chance of a female contracting measles is
, hence the probability of a person contracting
measles is
.
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
.
(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 . Hence the probability of
both cards
drawn are clubs is .
(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
.
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.
Let denote `the
first urn is chosen' , and denote
`the second urn is chosen' , and denote
`the third urn is
chosen' , and B denote `A ball drawn is red'.
According to BAYES'THEOREM, we have
And we know , and , and
, and , thus we get
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 .
There is a chance of having a storm on any
day, and the chance that no storm on any day is .
Hence the chance that there is no storm on three consecutive days
is .
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