If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

# Solved Examples: Probability Distribution Prerequisite Topics
Probability Distribution Calculators

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Show all work.

(1.) Samuel tossed a coin four times.
(a.) List the outcomes of his experiment.
(b.) List the random variable, $x$ for the number of tails in the outcomes.
(c.) Calculate the probability of the random variable, $P(x)$ for the outcomes.
(d.) Calculate the mean of the random variable.

We can use the Punnett Square or Tree Diagram to find the outcomes.
Let us use the Punnett Square.
You may choose to use Tree Diagram if you wish.
(a.)
2 coins on row, 2 coins on column
HH HT TH TT
HH HHHH HHHT HHTH HHTT
HT HTHH HTHT HTTH HTTT
TH THHH THHT THTH THTT
TT TTHH TTHT TTTH TTTT

(b.) and (c.)
$x$ $P(x)$ $x * P(x)$
$0$ $\dfrac{1}{16}$ $0$
$1$ $\dfrac{4}{16}$ $\dfrac{4}{16}$
$2$ $\dfrac{6}{16}$ $\dfrac{12}{16}$
$3$ $\dfrac{4}{16}$ $\dfrac{12}{16}$
$4$ $\dfrac{1}{16}$ $\dfrac{4}{16}$
$\Sigma x = 10$ $\Sigma[x * P(x)] = \dfrac{32}{16} = 2$

(d.)
$\mu = \Sigma[x * P(x)] \\[3ex] \mu = \dfrac{32}{16} \\[5ex] \mu = 2$
(2.) The table shows the random variable, $x$ and the probability of the random variable, $P(x)$
$x$ $0$ $1$ $2$ $3$
$P(x)$ $0.141$ $0.389$ $0.359$ $0.111$

The random variable represents the number of boys in a family with three children.
(a.) Calculate the mean of the random variable. Round to 2 decimal places.
(b.) Calculate the standard deviation of the random variable. Round to 2 decimal places.
(c.) Is it unusual for a family with three children to have only boys?

(a.)
$x$ $P(x)$ $x * P(x)$ $x^2$ $x^2 * P(x)$
$0$ $0.141$ $0$ $0$ $0$
$1$ $0.389$ $0.389$ $1$ $0.389$
$2$ $0.359$ $0.718$ $4$ $1.436$
$3$ $0.111$ $0.333$ $9$ $0.999$
$\Sigma[x * P(x)] = 1.44$ $\Sigma[x^2 * P(x)] = 2.824$

$\mu = \Sigma[x * P(x)] \\[3ex] \mu = 1.44 \\[3ex] (b.) \\[3ex] \sigma = \sqrt{\Sigma[x^2 * P(x)] - \mu^2} \\[3ex] \sigma = \sqrt{2.824 - 1.44^2} \\[3ex] \sigma = \sqrt{2.824 - 2.0736} \\[3ex] \sigma = \sqrt{0.7504} \\[3ex] \sigma = 0.8662563131 \\[3ex] \sigma \approx 0.87 \\[3ex] (c.) \\[3ex] \underline{First\;\;Method:\;\;Probability} \\[3ex] P(3) = 0.111 \\[3ex] 0.111 \gt 5\% \\[3ex] 0.111 \gt 0.05 \\[3ex]$ Therefore, it is not unusual for a family with three children to have only boys.

$\underline{Second\;\;Method:\;\;Usual\;\;Values} \\[3ex] Minimum\;\;usual\;\;value \\[3ex] = \mu - 2\sigma \\[3ex] = 1.44 - 2(0.8662563131) \\[3ex] = 1.44 - 1.732512626 \\[3ex] = -0.2925126262 \\[3ex] 3 \gt -0.2925126262 \\[3ex] Maximum\;\;usual\;\;value \\[3ex] = \mu - 2\sigma \\[3ex] = 1.44 + 2(0.8662563131) \\[3ex] = 1.44 + 1.732512626 \\[3ex] = 3.172512626 \\[3ex] 3 \lt 3.172512626 \\[3ex] -0.2925126262 \lt 3 \lt 3.172512626 \\[3ex]$ Therefore, it is not unusual for a family with three children to have only boys.
(3.) The table shows the results from groups of 10 births from 10 different sets of parents.

 $x$ $P(x)$ $0$ $0.003$ $1$ $0.016$ $2$ $0.038$ $3$ $0.113$ $4$ $0.209$ $5$ $0.235$ $6$ $0.203$ $7$ $0.116$ $8$ $0.043$ $9$ $0.013$ $10$ $0.011$

The random variable, $x$ represents the number of girls among 10 children.
(a.) Using the Range Rule of Thumb, determine the maximum usual value.
(b.) Using the Range Rule of Thumb, determine the minimum usual value.
(c.) Based on the result, is 1 girl in 10 births an unusually low number of girls? Explain.

$x$ $P(x)$ $x * P(x)$ $x^2$ $x^2 * P(x)$
$0$ $0.003$ $0$ $0$ $0$
$1$ $0.016$ $0.016$ $1$ $0.016$
$2$ $0.038$ $0.076$ $4$ $0.152$
$3$ $0.113$ $0.339$ $9$ $1.017$
$4$ $0.209$ $0.836$ $16$ $3.344$
$5$ $0.235$ $1.175$ $25$ $5.875$
$6$ $0.203$ $1.218$ $36$ $7.308$
$7$ $0.116$ $0.812$ $49$ $5.684$
$8$ $0.043$ $0.344$ $64$ $2.752$
$9$ $0.013$ $0.117$ $81$ $1.053$
$10$ $0.011$ $0.110$ $100$ $1.100$
$\Sigma[x * P(x)] = 5.043$ $\Sigma[x^2 * P(x)] = 28.301$

$\mu = \Sigma[x * P(x)] \\[3ex] \mu = 5.043 \\[3ex] \sigma = \sqrt{\Sigma[x^2 * P(x)] - \mu^2} \\[3ex] \sigma = \sqrt{28.301 - 5.043^2} \\[3ex] \sigma = \sqrt{28.301 - 25.431849} \\[3ex] \sigma = \sqrt{2.869151} \\[3ex] \sigma = 1.693856842 \\[3ex] (a.) \\[3ex] Maximum\;\;usual\;\;value \\[3ex] = \mu + 2\sigma \\[3ex] = 5.043 + 2(1.693856842) \\[3ex] = 5.043 + 3.387713683 \\[3ex] = 8.430713683 \\[3ex] (b.) \\[3ex] Minimum\;\;usual\;\;value \\[3ex] = \mu - 2\sigma \\[3ex] = 5.043 - 2(1.693856842) \\[3ex] = 5.043 - 3.387713683 \\[3ex] = 1.655286317 \\[3ex] (c.) \\[3ex] \underline{First\;\;Method:\;\;Probability} \\[3ex] P(1) = 0.016 \\[3ex] 0.016 \lt 5\% \\[3ex] 0.016 \lt 0.05 \\[3ex]$ Therefore, based on the result; 1 girl in 10 births is an unusually low number of girls.

$\underline{Second\;\;Method:\;\;Usual\;\;Values} \\[3ex] 1 \lt 1.655286317 \\[3ex]$ Therefore, based on the result; 1 girl in 10 births is an unusually low number of girls.
(4.) The table show the random variable, $x$ and the probability of the random variable, $P(x)$

 $x$ $P(x)$ $0$ $0.000$ $1$ $0.003$ $2$ $0.016$ $3$ $0.054$ $4$ $0.121$ $5$ $0.193$ $6$ $0.226$ $7$ $0.193$ $8$ $0.121$ $9$ $0.054$ $10$ $0.016$ $11$ $0.003$ $12$ $0.000$

The random variable represents the number of Hispanics on a panel of 12 jury in a predominantly Hispanic population.
(a.) Determine the probability of having exactly 5 Hispanics among 12 jurors.
(b.) Determine the probability of having 5 or fewer Hispanics among 12 jurors.
(c.) Is 5 Hispanics an unusually low number of Hispanics among 12 jurors for that population?

$(a.) \\[3ex] P(5) = 0.193 \\[3ex] (b.) \\[3ex] P(\le 5) = 0.000 + 0.003 + 0.016 + 0.054 + 0.121 + 0.193 \\[3ex] P(\le 5) = 0.387 \\[3ex] (c.) \\[3ex] 0.387 \gt 5\% \\[3ex] 0.387 \gt 0.05 \\[3ex]$ Hence, 5 Hispanics is not an unusually low number of Hispanics among 12 jurors for that population.
(5.)

(6.) Cardano plays lottery games.
If he picks a sequence of three digits from the numbers 0 through 9 in a lottery game, he will pay $1.35 because he has to pay$1.35 to play the game.
If he picks the same sequence of three digits in a drawing, he will win $487.75 (a.) How many different selections are possible? (b.) Determine the probability of winning the lottery game (c.) Calculate Cardano's net profit if he wins the lottery game (d.) Calculate the expected value (a.) The question did not specify whether the numbers can be picked: with or without repetition The default is to assume that ghe numbers can be picked with repetition. 0 through 9 consists of 10 digits$\underline{0 - 9}\underline{0 - 9}\underline{0 - 9}\underline{10}\underline{10}\underline{10}$Number of different selections =$10 * 10 * 10
= 1000$selections$ (b.) \\[3ex] n(Winning) = 1 * 1 * 1 = 1 \\[3ex] n(S) = 10 * 10 * 10 = 1000 \\[3ex] P(Winning) = \dfrac{n(Winning)}{n(S)} \\[5ex] = \dfrac{1}{1000} \\[5ex] (c.) \\[3ex] Net\;\;profit = Winning\;\;amount - Amount\;\;to\;\;play \\[3ex] = 487.75 - 1.35 \\[3ex] = \$486.40 \\[3ex]$ (d.)

$P(Winning) = \dfrac{1}{1000} \\[5ex] P(Losing) = 1 - P(Winning) ...Complementary\;\;Rule \\[3ex] P(Losing) \\[3ex] = 1 - \dfrac{1}{1000} \\[5ex] = \dfrac{1000}{1000} - \dfrac{1}{1000} \\[5ex] = \dfrac{1000 - 1}{1000} \\[5ex] = \dfrac{999}{1000} \\[5ex]$
Event $x$ $P(x)$ $x * P(x)$
Lose the game $-1.35$ $\dfrac{999}{1000}$ $-1.34865$
Win the game $486.40$ $\dfrac{1}{1000}$ $0.4864$
$\Sigma[x * P(x)] = -0.86225$

$E = \Sigma[x * P(x)] \\[3ex] E = -\$0.86225 \\[3ex] E \approx -\$0.86$
(7.)

(8.)

(9.)

(10.)