CS206 -Discrete Mathematics II Instructor: Chitoor V.Srinivasan PROBLEM SET 9 SOLUTIONS
Let n be the number of random digits and let X be the number of
occurrences of the digit 7. We want . Solving for
n yields , hence n =
22.
Let's consider the people to be indistinguishable (considering to
be distinct yields the same answer but much more difficultly). There are
ways to partition the 6 birthdays into 12 months.
To find the number which fall into 2 calendar months let's choose 2 months
which can be done in ways. For each such
choice of months,
their are non-empty partitions of
6 birthdays into the
two months. Thus the probability is
It's a Binomial distribution. . Solving for
we get and , which is
approximately zero. Note: some biologists claim that the chances of a
simple bacterium forming from the random collisions of atoms and molecules
is even much more remote than that!
It's Poisson distribution, with
.
It's Bernoulli trial, with .
(a).
It's Negative binomial distribution with r = 5 successes.
(b).
It's also Negative binomial distribution with r = 5 successes.
(c).
It's Binomial distribution, with n = 7.
This requires use of negative binomial distribution in which
``success'' is producing a defective item. The answer is the
probability that five or more tries are made before stopping, where
the stopping condition is, obtaining 3 successes (i.e. getting 3
defective items). This is,
The expected number of trials for the negative binomial distribution
is 3/(.15) = 20.
Lets define random variable Head to be
the no.of heads, and
Tail to be the no.of tails. These are both binomial
distributions
so and . Thus,
When and
.
So if the ticket is $5, it's better to bet, and if it's $10, it's
better not to bet.
Lets define the random value
to be the result of the
dice rolled first, lets define the random value to be the
result of the dice rolled second. We have
(a).
(b).
(c).
Lets assume the dice rolled first getting i and dice
rolled second getting j, and let
correspond to the entrance (i,j) of the following.
(d).
(e).
5s and 6s on 1st die)
5s on 1st die) 6s on 1st die)]
= 2 (1/6 + 1/6) = 2/3