Solving Genetics Probability Problems

In probability method, we have to decide the basic terms of probability.. The main aim of genetics probability subject is to give the fundamental properties the importance of their use with some basic examples of genetics. In a genetics probability, the basic term and properties is not easier to understand but work with sums we will start to get comfortable with in genetics probability.


Basic Definitions of genetics probability:

In a genetic probability method, an testing method is continuously repeated for infinite numbers of time and it happened by the expected probability in the genetics

Genetics probability solving problems:

Problem 1:

To solve consider a locus with two chromosome, C and c. If the frequency of CC is 0.64, what is the frequency of A under Hardy-Weinberg?

Solution:

Solving methods of genetic probability,

Under H-W, if q = freq(C) , then q2 = freq(CC), hence q2 =0.64 or q = 0.8.

Problem 2:

To solve if the genotypes CC, Cc, and cc have frequencies 0.7, 0.49, and 0.49 (respectively), what are q = freq (C)? r = freq (c)? After a single generation of random mating, what is the expected frequency of CC? Of Cc? Of cc?

Solution:

Solving methods of genetic probability,

p = freq (CC) + (1/2) freq (Cc) = 0.7 + (1/2)(0.49) = 0.945

q = 1-p = 0.945

freq (CC) = p2 = 0.9452 = 0.8930

freq (Cc) = 2qr = 2*0.945*0.8930 =1.6877

freq (cc) = r2= 0.89302= 0.797449

Problem 3:

To solve the suppose 60 out of 850 women’s are green eyes. What is frequency of green eyes? If a random women’s is chosen, what is the probability they are a green eyes?

Solution:

Solving methods of genetic probability,

Freq (Redheads) = 60/850 = 0.070 or 7.05 percent

Probability of a Redhead = 30/850, or 7.05 percent

Problem 4:

Solve the genotype cc is lethal and yet population has an equilibrium frequency for c of 30.  Here the Cc is the fitness and its value is one.  Find the CC genotype fitness?

Recall if the genotypes CC: Cc: cc have fitness 1-s : 1 : 1-t, then the equilibrium frequency of A is s/(r+s).

Here, s= 1 (as cc is lethal), so that freq(C) = 1- freq(c) =0.6 = 1/(1+s)

Solving gives 1+s = 1/0.8, or s = 1/0.8 -1 = 3.5

Hence fitness of CC = 1-s = 1-3.5=-2.5.

Learn Derivative at a Point

Learn derivative at a point involves the process solving derivative problems at an exam point view. The rate of change of the given function is determined with the help of calculus. To find the rate of change the calculus is divided into two types such as differential calculus and integral calculus. At exam point of view the derivative is easily carried out by differentiating the given function with respect to input variable. The following are the solved example problems in derivatives at exam point of view for learn.

Learn derivative at a point example problems:


Example 1:

Find the derivative for the given differential function.

f(e) = 3e 3 -4 e 4  - 5e

Solution:

The given function is

f(e) = 3e 3 -4 e 4  - 5e

The above function is differentiated with respect to e to find the derivative

f '(e) = 3(3e 2 )-4(4e 3  ) - 5

By solving above terms

f '(e) = 9e 2 -  8e 3 - 5

Example 2:

Compute the derivative for the given differential function.

f(e) = 6e6 - 5 e5 - 4 e4 - e

Solution:

The given equation is

f(e) = 6e6 - 5 e5 - 4 e4 - e

The above function is differentiated with respect to e to find the derivative

f '(e) =  6(6e 5)  -5 (5 e4 ) -4(4 e3) - 1

By solving above terms

f '(e) =  36e 5  - 25 e4  - 16 e3 - 1

Example 3:

Compute the derivative for the given differential function.

f(e) = 2e 2 -4 e 4  - 15

Solution:

The given function is

f(e) = 2e 2 -4e 4  - 15

The above function is differentiated with respect to e to find the derivative

f '(e) = 2(2e  )-4(4 e 3 ) - 0

By solving above terms

f '(e) = 4e -16e3

Example 4:

Compute the derivative for the given differential function.

f(e) = 5e5 -4e 4 -3e 3  - 2

Solution:

The given function is

f(e) = 5e5 -4e 4 -3e 3  - 2

The above function is differentiated with respect to e to find the derivative

f '(e) = 5(5e 4 )-4(4e 3 ) -3( 3e 2) -0

By solving above terms

f '(e) = 25e 4 -16e 3  -9 e 2


Learn derivative at a point practice problems:


1) Compute the derivative for the given differential function.

f(e) = 2e 3 -3 e 4  - 4 e 5

Answer: f '(e) = 6e 2 -12 e3 - 20 e 4

2) Compute the derivative for the given differential function.

f(e) = e 3-e5 - 4 e 6

Answer: f '(e) = 3e2 - 5e4 - 24 e 5

Linear Algebra Test

Linear function is a function with polynomial degree 1. This linear function matches a dependent variable with independent variable and maintains a relation in a simpler way. The linear algebra is the important method which is used in the linear function, variable function, linear equation, vector, matrix etc. This linear function is used in the high school and college students are using the linear function by using the linear algebra.

Sample Problem for linear algebra:


f is a linear function. Value of x and f(x) are given in the table below; complete the table using linear algebra

X                           f(x)

-1                          11

1                           –

–                           1

6                          -12

7                         --

–                          -35

Solution:

f is a linear function in linear algebra whose formula has the form to test is

f(x) = a x + b

where a and b are constants to be test and found. Note that 2 ordered pairs (-2,18) and (5,-19) are given in the table. These are two ordered pairs to test are used to write a system of linear equations as follows

11 = - 1 a + b and -12 = 6 a + b

Solve the above system to obtain a = - 3 and b = 2 and write the formula for function f as follows

f(x) = - 3 x + 2

We now use the formula for the f to find f(x) given x or find x given f(x).

for x = 0 , f(0) = -3(0) + 2 = 2

for f(x) = 1 , 1 = -3 x + 2 which gives x = 1/3

for x = 7 , f(7) = -3(7) + 2 = - 19

for f(x) = - 30 , -30 = -3 x + 2 which gives x = 32/3

We now put the values of the calculated above in the answer.

X                           f(x)

-1                          11

1                           2

1/3                         1

6                         -12

7                         -19

32/3                      -35


Practice problem:


Problem 1:
f is a linear function in linear algebra. Value of x and f(x) to test are given in the table below; complete the table.

X                              f(x)

-5                             15

1                               –

–                              1

10                         -14

8                             --

–                           -35


Problem 2:
f is a linear function. Value of x and f(x) to test are given in the table below; complete the table.

X                               f(x)

-6                             18

1                              –

–                             1

9                           -20

10                             --

–                             -40

Exponentiation Tutoring

This article is a online exponentiation tutoring.

Generally the exponentiation is used in all sciences including the mathematical, physical, chemical or even in biological sciences but it is assumed or preferably called a mathematical operation which generally written as an it is clear from this notation that the exponentiation involves two numbers one is called the base i.e. a and other is called the exponent i.e. n. When n is any positive integer then the exponentiation simply means the repeated multiplication of the number ‘a’ to itself n times which are written as:

 

Generally, exponent is the superscript to the base. Exponentiation above will be read as a raised to the power n or exponent of n, or a to the n. A few exponents have their pronunciation: for instance, a2 is read as a square and a3 as a cube.

Exponentiation comes to use in economics or biology or chemistry or physics or computer science to calculate compound interest or population growth.


Example problems on exponentiation tutoring:


Positive integer exponents:

a2 is pronounced as a square because any square with side a has area equal to a2. The expression a3 is referred to as the cube of a since the volume of a cube with side length a is a3.

So 32 is called as "three squared", and 23 is "two cubed".

The numbers with positive integer exponents may be defined by the initial condition

a1 = a

thus in general the recurrence relation

an+1 = a·an.


Exponents of one and zero:


If a is not zero and n an m are two positive integer exponents such that the exponent (n – m) is also positive then we can say that



If we consider a special case in which the exponent’s n and m are equal then we can wrote the equality as

 

Thus we lead to the following rule: Any number with raised to the power 1 is the number itself.

Solving Quadratic Formula

Quadratic Equation:

An equation which has one or more terms are squared but no higher power in terms, having the syntax, ax2+bx+c =0 where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant numerical term.

Types of quadratic equation

Pure quadratic equation:

The numerical coefficient cannot be zero. If b=0 then the quadratic equation is called as a ‘pure’ quadratic equation

Complete quadratic equation:

If the equation having x and x2 terms such an equation is called a ‘complete’ quadratic equation. The constant numerical term ‘c’ may or may not be zero in a complete quadratic equation. Example, x2 + 5x + 6 = 0 and 2x2 - 5x = 0 are complete quadratic equations.


Quadratic Equation Formula


The quadratic equation has the solutions ax2+bx+c =0

x =√(b2-4ac)/2a

Consider the general quadratic equation

ax2+bx+c =0

With a`!=` 0. First divide both sides of the equation by a to get

x2+b/a x + c/a =0

which leads to

x2+ b/a x = - c/a

Next complete the square by adding ((b)/(2a) )2to both sides

X2+ ((b)/(a) )x+((b)/(2a) )2 = -((c)/(a) )+ ((b)/(2a) )2

(x+(b)/(2a) )2=-((c)/(a) ) + ((b^2)/(4a^2) )

(x+(b)/(2a) )2 = (b^2-4ac)/(4a^2)

Finally we take the square root of both sides:

x+(b)/(2a) = +-(sqrt(b^2-4ac))/(2a)

or

x =-(b)/(2a)  +-(sqrt(b^2-4ac))/(2a)

The final form of Quadratic Formula is

x =-b+-sqrt(b^2-4ac)/(2a)

The two roots of the equation is

``          -b-(sqrt(b^2-4ac))/(2a)

-b+(sqrt(b^2-4ac))/(2a)


Example Problem on solving Quadratic formula


Example:

Find the roots of the equation by quadratic formula method, x2-10x+25=0

Solution:

Step 1:  From the equation, a = 1, b = - 10 and c = 25.

Step 2:  To Find X:
plug-in the values in the formula below
x = ``

Step 3:  We get the roots, x = ``
x = 5 and x = 5
which means x1 = 5 and x2 = 5.

Here x = 5 is root of the equation.

learning problems in probability

Introduction:

Learning Probability is also one part of mathematics subject. Probability disturbed with investigation of approximate phenomena. Learning the fundamental things of the probability is random variables, stochastic processes, and events. Mathematical abstraction of the non-deterministic events or calculated quantities that may moreover be single occurrences or evolve over time in an apparently random fashion. While a coin toss or roll of a die is a approximate event.

Let us some example problems in probability with solved solutions.

Learning example problems in probability:

Learning Example problem 1:

Three dice are rolled once. What is the chance that the sum of the expression numbers on the three dice is greater than 15?

Solution:

In probability problems while three dice are rolled,

the trial space S = {(1,1,1), (1,1,2), (1,1,3) ...(6,6,6)}.

S contains 6 × 6 × 6 = 216 outcomes.

Let A be the event of receiving the sum of appearance numbers greater than 15.

A = { (4,6,6), (6,4,6), (6,6,4), (5,5,6), (5,6,5), (6,5,5), (5,6,6), (6,5,6),

(6,6,5), (6,6,6)}.

N (S) = 216 and n (A) = 10.

Therefore P (A) equal to n (A) / n(S)

= 10/ 216

=5 / 108.


Probability Problem 2:


Learning Example problem 2:

In a vehicle Stand there are 100 vehicles, 60 of which are cars, 30 are mini truck and the rest are Lorries. If each vehicle is uniformly likely to leave, find out the probability in this problem.

a) Mini truck departure first.

b) Lorry departure first.

c) Car leaving second if either a lorry or mini truck had gone first.

Solution:

a) Let S be the model space and A be the event of a mini truck leaving first.

n (S) = 100

n (A) = 30

Probability of a mini truck departure first:

P (A) = 30 / 100 = 3 / 10.

b) Let B be the event of a lorry departure first.

n (B) = 100 – 60 – 30 = 10

Probability of a lorry departure first:

P (B) = 10 / 100 = 1 / 10.

c) If moreover a lorry or mini truck had gone first, then there would be 99 vehicles totally left there, 60 of which are cars. Let T be the trial space and C be the occurrence of a car departure.

n(T) = 99

n(C) = 60

Probability of a car departure after a lorry or mini truck has gone:

P(C) = 60 / 99 = 20 / 33.

Learning Probability Density Functions

The Learning Probability density functions are known as Gaussian functions. A probability density function of a continues random variable is defined as a function that describes the virtual probability for that random variable to occur at a given point within the using space. The Learning of probability density functions is same for the probability distribution functions. In this article we see the learning  the definition of probability density functions and some example problem for the probability density functions.

Definition of Learning the Probability Density Functions:


Probability density function is defined as continues random variable functions f(x). this functions satisfies the following properties.

Probability functions limits between a and b

P(a≤x≤b) = ∫ a to b  f(x) . dx

A probability density function has only real for the real value.

F(x) ≥ 0

Integral of the probability density functions is 1.

∫ -oo to oo f(x) dx  = 1

Learning the fundamental properties of probability density functions:

F(x) is continues random functions

P(a<= x <b) =  P(a < x<b) = P( a< x <= b)

= P(a<=X<b) =

Where f is probability density distribution functions

In the differentional functional of the probability density functions, we have

P(x<X<=x+dx) = F(x+dx)-F(x) = dF(x)=F'(x)dx = f(x)dx

Where

X is Known as probability differential functions.


Learning the some example problems for the probability density functions:


Example 1:

If the probability density function of a random variable is given by f(x) = H (x – x^3);  0 < x <1 .find the H.

Solution:

Since ,

F(x) is the probability density functions means,

F(x)= ∫ a to b f(x) dx

Substitute the values,

H [ x^2/2 – x^4/4] limits 0 to1 = 1

H [ 1/2- 1/4] = 1

1 After simplify, We get

H= 4

Example-2

If the probability density function of a random variable is given by f(x) = H (1– x^5);  0 < x <1 .find the H.

Solution:

Since ,

F(x) is the probability density functions means,

F(x)= ∫ a to b f(x) dx

Substitute the values,

H [ x – x^6/6] limits 0 to1 = 1

H [ 1- 1/6] = 1

1 After simplify, We get

H= 6/5 = 1.2

Practice problem of  learning the probability density functions:

Problem -1:

A continuous random variable X follows the probability law, f(x) =H x (x – x^2 ), 0 < x < 1and 0 for elsewhere. Find k

Answer: H = 12

Problem -2:

A continuous random variable X follows the probability law, f(x) =H x^2 (x– x^4 ), 0 < x < 1and 0 for elsewhere. Find k

Answer: H = 9.33