# Probability Distribution Homework Help

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Probability is a way of determining how likely something is to happen, and the area of math called probability theory gives us the mathematical tools necessary to analyze each situation and perform the necessary calculations. Probability is used extensively in statistics, finance, gambling, science, and philosophy.

A rigorous course in probability will cover many of the topics listed below:

• The need for measure theory
• Probability triples
• Probabilistic foundations
• Expected values
• Inequalities and convergence
• Distributions of random variables
• Stochastic processes and gambling games
• Discrete Markov chains
• Probability theorems
• Weak convergence
• Characteristic functions
• Decomposition of probability laws
• Conditional probability and expectation
• Martingales
• General stochastic processes

A large number of good books on probability can be found; a good place to start being Amazon.com. In addition, there is an excellent probability tutorial at West Texas A&M University.

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## Topics Covered in Probability Homework Help

Given below are some of the important topics of probability covered by our homework help:

• Complimentary Events
• Conditional Probability
• Compound Independent Events
• Dependent Events
• Mutually Exclusive Events
• Theoretical Probability

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Question: Consider families with $4$ children each. What percentage of families would you expect to have

(a) two boys and two girls

(b) at least one boy

(c) at most two boys

(d) no boys

(Assume probability of getting a girl and probability of getting a boy are equal and is $\frac{1}{2}$)

Solution:

Step 1:

Since there are $4$ children $(n)$ = $4$.

The event of getting boy is a success and event of getting girl is a failure. The probability of getting a boy and getting a girl is equally likely.

So probability of getting a boy = Probability of getting a girl = $\frac{1}{2}$

That is $p$ = $q$ = $\frac{1}{2}$

Step 2:

Binomial Distribution formula is given by

$P(x)$ = $(nCx)\ p^x\ q^{n-x}$, where $x$ = $0, 1, 2, …...,n$

Now lets consider each case.

Let $x$ be the number of boys

(a) Two boys and two girls

$\Rightarrow\ x$ = $2$.

Substituting $x$ = $2$ in the formula

$P$(Two boys and two girls) = $p(x = 2)$ = $4c_{2}\left ( \frac{1}{2} \right )^2\left ( \frac{1}{2} \right )^{4-2}$ = $4c_2\left ( \frac{1}{2} \right )^2\left ( \frac{1}{2} \right )^2$

(b) at least one boy

=> x = 1, 2, 3 or 4.

P(at least one boy) = $p(x = 1) + p(x = 2) + p(x = 3) + p(x = 4)$

= $4c_1\left ( \frac{1}{2} \right )^1\left ( \frac{1}{2} \right )^{4-1}+ 4c_2\left ( \frac{1}{2} \right )^2\left ( \frac{1}{2} \right )^{4-2}+4c_3\left ( \frac{1}{2} \right )^3\left ( \frac{1}{2} \right )^{4-3}+4c_4\left ( \frac{1}{2} \right )^4\left ( \frac{1}{2} \right )^{4-4}$

= $4c_1\left ( \frac{1}{2} \right )^1\left ( \frac{1}{2} \right )^3+4c_2\left ( \frac{1}{2} \right )^2\left ( \frac{1}{2} \right )^{2}+4c_3\left ( \frac{1}{2} \right )^3\left ( \frac{1}{2} \right )^1+4c_4\left ( \frac{1}{2} \right )^4\left ( \frac{1}{2} \right )^0$

= $\frac{4}{16}+\frac{6}{16}+\frac{4}{16}+\frac{1}{16}$

= $\frac{15}{16}$

c) at most two boy

=> x = 0, 1 or 2.

P(at most two boy) = $p(x = 0) + p(x = 1) + p(x = 2)$

= $4c_0\left ( \frac{1}{2} \right )^0\left ( \frac{1}{2} \right )^{4-0}+4c_1\left ( \frac{1}{2} \right )^1\left ( \frac{1}{2} \right )^{4-1}+4c_2\left ( \frac{1}{2} \right )^2\left ( \frac{1}{2} \right )^{4-2}$

= $4c_0\left ( \frac{1}{2} \right )^0\left ( \frac{1}{2} \right )^4+4c_1\left ( \frac{1}{2} \right )^1\left ( \frac{1}{2} \right )^3+4c_2\left ( \frac{1}{2} \right )^2\left ( \frac{1}{2} \right )^2$

= $\frac{1}{16}+\frac{4}{16}+\frac{6}{16}$

= $\frac{11}{16}$

d) No boys

=> x = 0

Substituting x = 0 in the formula

P(No boys) = p(x = 0)

= $4c_0\left ( \frac{1}{2} \right )^0\left ( \frac{1}{2} \right )^{4-0}$

=$4c_0\left ( \frac{1}{2} \right )^0\left ( \frac{1}{2} \right )^4$

= $\frac{1}{16}$