# Math Homework Help Free Probability Worksheet

## Probability Worksheets

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## Detailed Description for All Probability Worksheets

**Definitions of Probability Worksheet**

These *Probability Worksheets* are a great handout for reinforcing the definitions of Probability.

**Probability on Numbers Worksheet**

These *Probability Worksheets* will produce problems with simple numbers between 1 and 50.

**Probability With a Single Die Worksheet**

These *Probability Worksheets* will produce problems with simple numbers, multiples, divisors, and factors using a single die.

**Probability With a Pair of Dice Worksheet**

These *Probability Worksheets* will produce problems with simple numbers, sums, differences, multiples, divisors, and factors using a pair of dice.

**Probability With a Deck of Cards Worksheet**

These *Probability Worksheets* will produce problems about a standard 52 card deck without the Jokers.

**Probability Using a Spinner Worksheet**

These *Probability Worksheets* will produce problems using a spinner.

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## Topics Covered in Probability Homework Help

Back to TopGiven 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 $x$ = $0, 1, 2, â€¦...,n$

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)

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

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

= $\frac{1}{16}$

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