Note: If you assumed that the probability of selecting
the number 0.345 was 1/1000, you got what appears to
be the correct answer using the wrong probability.
5.1 Problem 28
Suppose that a message consisting of six characters
is transmitted. If each character consists of seven
bits (plus a parity check bit), what is the probability
that a message is erroneously recieved, but none of
the errors is detected by the parity check?
Solution: We are given the probability that each bit
that is sent is correctly is recieved correctly with
a probability of 0.999, independently of the other
bits. An undetected error in a single character occurs
when 2, 4, 6, or 8, of the bits are recieved in error.
The probability of this occuring is given by:
Let
be the probability that a single character has errors
but is undetected. If X is the number of the six characters
that contain undetected errors, then X is a binomial
random variable with parameters 6 and
. The probabilty of getting 1 or more undetected errors
is found by calculating:
5.2 Problem 14
In a certain town, crimes occur at a Poisson rate of
five per month. What is the probability of having exactly
two months (not neceassily consecutive) with no crimes
during the year?
Solution: First we find the probability of no crimes
in a single month. We select a month as the unit of
time. Then
and the probability of no crimes during any given month
of a year is
.
If X is the number of times that there is no crime during
a month, X is a binomial random variable and the probability
that X = 2 , (no crime, in any two of the 12 months)
is:
5.2 Problem 20
Suppose that during a year 94% of the letter carriers
are not bitten by dogs. Assuming that dogs bite letter
carriers randomly, what percentage of those who sustained
one bite will be bitten again?
Solution: Let N be the total number of letter carriers
for a given year. Let X be the count of the number
of bites a letter carrier has per year. Assuming X
is a random Poisson variable we can determine the parameter
by calculating:
from which we get:
= 0.06187540371
Next, we calculate the conditional probability
P( X > 0 |
) =
.
So, about 3% of the letter carriers will be bitten more
than once.
Chapter 5.1 - 5.2 HW Turn-in
Solutions
5.1 Problem 6
A manufacturer of nails claims that only 3% of its nails
are defective. A random sample of 24 nails is selected,
and two of them are defective. Is it fair to reject
the manufacturer's claim based on this observation?
Solution: First, we calculate the probability of getting
a random sample with two or more defective nails using
the manufacturer's claim of 3% defective..
Let X be the number of defective nails. Then X is a
binomial random varible with:
and with the probability function:
The probability of getting two or more nails in a random
sample of 24 nails is:
or roughly once in every six samples. (Compare this
with the probability that if 4% of the nails are defective,
once in every four samples you will get 2 or more defective
nails.) Two or more defective nails in every six samples
occurs too frequently to allow me to reject the manufacturer's
claim given only one sample of 24 nails.
5.1 Problem 14
From the set {
:
}, 100 independent numbers are selected at random and
rounded to three decimal places. What is the probability
that at least one of them is 0.345?
Solution: In this problem, the sample space S is:
S = { 0.000, 0.001, 0.002, ... , 0.999, 1.000 }
There are 1001 elements in S
not
equiprobable. The probability of selecting a number
x that will be rounded to 0.345 is P(
), the length of this interval is 0.000999..., so by
the definition of randomly selected point from an interval
(p. 33 text),
P(
) = 0.000999/1 = 0.000999
Let Y be the number of times that we randomly select
0.345 in 100 tries. Then Y is a binomial random variable
with parameters:
