# Math Homework Help Free Probability Worksheet

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## Dynamically Created Probability Worksheets

These dynamically created Probability Worksheets are great for learning and practicing the concept of probability. These Probability Worksheets are ideal for 4th Grade, 5th Grade, 6th Grade, and 7th Grade students.

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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

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• 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)

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

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

= $\frac{1}{16}$