- Since we have
And if we can show that
Then we can finish our proof. Now let's use mathematics induction to
prove the above formula.
Basis:
Induction Hypothesis: Let's assume following holds.
Induction Step: If we can show
then it's all done.
According to Theorem 2.3.6, we have
- Since A and B are events
defined over S ,hence we
have
. And
Substituting
, we get
- The sample outcomes which the sum of the two fair dices exceeds
8 are following
The sample outcomes which the sum of the two fair dices equals 10
are following
Hence the probability that the sum equals 10 given it exceeds
8 is 3/10.
- Let B denote that the chip
drawn from urn II is red, and
denote that the chip drawn from urn I and
transferred to urn II
is red, and
denote that the chip
drawn from urn I and
transferred to urn II is red, hence we get
- Poisson Distribution
,
where
is the mean value.
- The chance of a point selected randomly locates inside the
triangle is the ratio of the area of triangle, which is
, to the area of circle, which is
.
And we consider it that we choose points randomly as Bernolli Trials,
hence we get the chance of a point located inside the triangle, p,
which is
, and the chance of a
point located
outside the triangle, q, which is
.
- By the multinomial theorem.
Since the only way to use
to construct
13 is to sum
ninty-seven 0, two 4 and one 5. Thus the
coefficient of
is
.
- Lets assume the length
of the series of random digits is
n, and we have
ten digits.
Probability that a particular digit is neither 5 nor 7 is 8/10. Hence
probability of getting at least one 5/7 is
. Solve
to get the minimum value of n required.
- We have totally
32 cards, which are from 2 to 8.
There are
ways to draw five
cards from the thirty-two
cards. - There are totally
random function from
the domain
to the range
, since each
item in
the domain will be able to map m items in the range. If
the
function would have exactly k elements in its range, it
means
the random function is an onto function. Besides we will have
ways to choose the range of the onto function. For
a
spcific range of k elements, the number of onto function is
Hence the total number of onto function from the domain
to the range of k elements is
Thus the probability that the onto function would have exactly
k
elements in its range is