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
Students will:
(1.) Compute the conditional probabilities of events.
(2.) Apply the Fundamental Counting Principle.
(3.) Determine the number of permutations of items.
(4.) Determine the number of permutations of total items taking some items at a time.
(5.) Determine the number of combinations of total items taking some items at a time.
(6.) Solve applied problems on combinatorics.
NOTE: Students may also use technology to solve the questions on Combinatorics.
To use Technology to solve Combinatorics problems, please view the video
Answer all questions.
Show all work.
Wherever applicable, use at least two methods (two or more methods).
Students can decide whichever method they prefer.
(1.) $A$ and $B$ are events.
$
P(A\:\:AND\:\:B) = 0.4 \\[3ex]
P(A) = 0.5 \\[3ex]
$
Determine $P(B | A)$
Type an integer or decimal.
$
AND \implies \cap \\[3ex]
P(B | A) = \dfrac{P(B \cap A)}{P(A)} \\[5ex]
P(B \cap A) = P(A \cap B) = 0.4 \\[3ex]
P(A) = 0.5 \\[3ex]
\rightarrow P(B | A) = \dfrac{0.4}{0.5} \\[5ex]
P(B | A) = 0.8
$
(2.) $C$ and $D$ are events.
$
n(C\:\:AND\:\:D) = 430 \\[3ex]
n(C) = 830 \\[3ex]
$
Determine $P(D | C)$
Round to three decimal places as needed.
$
AND \implies \cap \\[3ex]
P(D | C) = \dfrac{n(D \cap C)}{n(C)} \\[5ex]
n(D \cap C) = n(C \cap D) = 430 \\[3ex]
n(C) = 830 \\[3ex]
\rightarrow P(D | C) = \dfrac{430}{830} \\[5ex]
P(D | C) = 0.518072289 \\[3ex]
P(D | C) \approx 0.518
$
(3.) A card is selected from a standard deck of cards.
(a.) Determine the probability that the card is a Queen.
(b.) Determine the probability that the card is a Queen given that the card is a Court.
Round your answers to three decimal places as needed.
(4.) According to the National Center for Education Statistics of the United States Department of
Education, the status dropout rate represents the percentage of youths (percentage of $16-$ to $24-year$ olds who
are not enrolled in school and have not earned a high school credential (either a diploma or an
equivalency credential such as a GED certificate).
In $2017$, the overall status dropout rate was $5.4\%$
Of that status dropout rate, Asian youths had the lowest status dropout rate of $2.1\%$
Determine the probability that a randomly selected dropout is an Asian given that the
person is between $16-24$ years old?
Round to four decimal places as needed.
(5.) Pride Boys Polls wanted to run a patriotic test of Americans.
The survey determined the responses of several age groups on how likely they were to purchase Made in America products.
The results of the survey are:
Likelihood of Purchase
$18-34$
$35-44$
$45-54$
$55+$
Total
More likely
$212$
$347$
$391$
$407$
$\boldsymbol{1357}$
Less likely
$20$
$7$
$22$
$13$
$\boldsymbol{62}$
Neither more nor less likely
$298$
$208$
$152$
$118$
$\boldsymbol{776}$
Total
$\boldsymbol{530}$
$\boldsymbol{562}$
$\boldsymbol{565}$
$\boldsymbol{538}$
$\boldsymbol{2195}$
(a.) One of the participants is selected at random.
Determine the probability that the person is at least $55$ years old, given that the
individual is less likely to buy a Made in America product.
(b.) One of the participants is selected at random.
Determine the probability that the individual is less likely to buy a Made in America product given that the person is at least $55$ years old.
(c.) Are $18-to-34-year$ olds more likely to buy Made in America products than individuals in
general? Give reasons for your answer.
Round all applicable answers to three decimal places as needed.
At least $55$ means $55$ or more
It means $55+$
$
Let: \\[3ex]
More\:\:likely = M \\[3ex]
Less\:\:likely = L \\[3ex]
55+ = A \\[3ex]
n(A) = 538 \\[3ex]
n(L) = 62 \\[3ex]
n(A \cap L) = 13 \\[3ex]
(a.) \\[3ex]
P(A | L) = \dfrac{n(A \cap L)}{n(L)} \\[5ex]
P(A | L) = \dfrac{13}{62} \\[5ex]
P(A | L) = 0.209677419 \\[3ex]
P(A | L) \approx 0.210 \\[3ex]
(b.) \\[3ex]
P(L | A) = \dfrac{n(L \cap A)}{n(A)} \\[5ex]
P(L | A) = \dfrac{13}{538} \\[5ex]
P(A | L) = 0.024163569 \\[3ex]
P(A | L) \approx 0.024 \\[3ex]
$
(c.)
Based on the survey:
$212$ people between $18-34-years$ are more likely to buy Made in America products.
$347$ people between $35-44-years$ are more likely to buy Made in America products.
$391$ people between $45-54-years$ are more likely to buy Made in America products.
$407$ people aged $55\:years$ and above are more likely to buy Made in America products.
$212 \lt 347 \lt 391 \lt 407$
This means that $18-34$ are not more likely to buy Made in America products.
(6.) Ruth received a shipment of eight laptops.
Three of the laptops are defective.
Two laptops are randomly selected.
(a.) Determine the probability that both laptops work.
(b.) Determine the probability that at least one of the laptops does not work.
The first method is the use of Multiplication and Addition Rules
At least $1$ means $1$ or more
If at least $1$ does not work, this means:
the first laptop is defective AND the second laptop works OR
the first laptop works AND the second laptop is defective OR
Both laptops are defective
The second method is the use of the Complementary Rule Event: At least one laptop is defective Complement of the Event: Both laptops are NOT defective (Both laptops work)
$
\underline{First\:\:Method} \\[3ex]
P(at\:\:least\:\:D) \\[3ex]
= P(DW) \\[3ex]
OR \\[3ex]
P(WD) \\[3ex]
OR \\[3ex]
P(DD)...Addition\:\:Rule \\[3ex]
Two\:\:laptops\:\:are\:\:selected \\[3ex]
This\:\:means\:\:Dependent\:\:Events \\[3ex]
This\:\:means\:\:Without\:\:Replacement\:\:condition \\[3ex]
P(DW) = \dfrac{3}{8} * \dfrac{5}{7} = \dfrac{15}{56} \\[5ex]
P(WD) = \dfrac{5}{8} * \dfrac{3}{7} = \dfrac{15}{56} \\[5ex]
P(DD) = \dfrac{3}{8} * \dfrac{2}{7} = \dfrac{6}{56} \\[5ex]
P(at\:\:least\:\:D) = P(DW) + P(WD) + P(DD) \\[3ex]
P(at\:\:least\:\:D) = \dfrac{15}{56} + \dfrac{15}{56} + \dfrac{6}{56} \\[5ex]
P(at\:\:least\:\:D) = \dfrac{15 + 15 + 6}{56} \\[5ex]
P(at\:\:least\:\:D) = \dfrac{36}{56} = \dfrac{9}{14} \\[5ex]
P(at\:\:least\:\:D) = 0.642857143 \\[3ex]
P(at\:\:least\:\:D) \approx 0.643 \\[3ex]
\underline{Second\:\:Method - Complementary\:\:Rule} \\[3ex]
P(at\:\:least\:\:D) + P(both\:\:work) = 1 ...Complementary\:\:Rule \\[3ex]
P(at\:\:least\:\:D) = 1 - P(both\:\:work) \\[3ex]
P(at\:\:least\:\:D) = 1 - P(WW) \\[3ex]
P(at\:\:least\:\:D) = 1 - \dfrac{5}{14} \\[5ex]
P(at\:\:least\:\:D) = \dfrac{14}{14} - \dfrac{5}{14} \\[5ex]
P(at\:\:least\:\:D) = \dfrac{14 - 5}{14} \\[5ex]
P(at\:\:least\:\:D) = \dfrac{9}{14} \\[5ex]
P(at\:\:least\:\:D) = 0.642857143 \\[3ex]
P(at\:\:least\:\:D) \approx 0.643 \\[3ex]
$
(a.) The probability that both laptops work is $0.357$
(b.) The probability that at least one laptop does not work is $0.643$
(7.) Cecilia bought a compact disk (CD) that has nine tracks.
After listening to the CD, she liked two of the songs.
With the random feature on her CD player, each of the nine songs is played once in random order.
Determine the probability that among the first two songs that was played,
(a.) She liked both of them.
(b.) Would this be unusual?
(c.) She liked neither of them.
(d.) She liked exactly one of them.
A song can be replayed before all nine songs are played.
Determine the probability that among the first two songs that was played,
(e.) She liked both of them.
(f.) Would this be unusual?
(g.) She liked neither of them.
(h.) She liked exactly one of them.
Round all applicable answers to three decimal places as applicable.
$
Let: \\[3ex]
Like\:\:the\:\:song = L \\[3ex]
Did\:\:not\:\:like\:\:the\:\:song = L' \\[3ex]
n(S) = 9 \\[3ex]
n(L) = 2 \\[3ex]
n(L') = 9 - 2 = 7 \\[3ex]
P(L) = \dfrac{n(L)}{n(S)} \\[5ex]
P(L) = \dfrac{2}{9} \\[5ex]
P(L') = \dfrac{n(L')}{n(S)} \\[5ex]
P(L') = \dfrac{7}{9} \\[5ex]
Two\:\:songs\:\:are\:\:selected \\[3ex]
This\:\:means\:\:Dependent\:\:Events \\[3ex]
This\:\:means\:\:Without\:\:Replacement\:\:condition \\[3ex]
(a.) \\[3ex]
P(LL) = \dfrac{2}{9} * \dfrac{1}{8} \\[5ex]
P(LL) = \dfrac{2}{72} \\[5ex]
P(LL) = \dfrac{1}{36} \\[5ex]
P(LL) = 0.027777778 \\[3ex]
P(LL) \approx 0.028 \\[3ex]
(b.) \\[3ex]
0.028 \lt 0.05 \\[3ex]
$
Yes, it is unusual that she liked the first two songs that was played because the probability that she liked both songs is less than
$5\%$ ... Usual and Unusual Events
$
(c.) \\[3ex]
P(L'L') = \dfrac{7}{9} * \dfrac{6}{8} \\[5ex]
P(L'L') = \dfrac{7}{3} * \dfrac{1}{4} \\[5ex]
P(L'L') = \dfrac{7}{12} \\[5ex]
P(L'L') = 0.583333333 \\[3ex]
P(L'L') \approx 0.583 \\[3ex]
$
(d.)
We can solve this question in two ways.
Use any method you prefer.
The first method is the use of Multiplication and Addition Rules
Liking exactly $1$ of them means:
She liked the first song AND did not like the second song OR
She did not like the first song AND liked the second song
The second method is to consider all cases/scenarios and sum the probabilities to one
The probability that she liked both songs + the probability that she liked neither song + the probability that she liked exactly one song = $1$
$
\underline{First\:\:Method} \\[3ex]
P(exactly\:\:L) P(LL')\:\:OR\:\:P(L'L) \\[3ex]
P(exactly\:\:L) P(LL') + P(L'L) ...Addition\:\:Rule \\[3ex]
Two\:\:songs\:\:are\:\:selected \\[3ex]
This\:\:means\:\:Dependent\:\:Events \\[3ex]
This\:\:means\:\:Without\:\:Replacement\:\:condition \\[3ex]
P(LL') = \dfrac{2}{9} * \dfrac{7}{8} = \dfrac{14}{72} ...Multiplication\:\:Rule\:\:for\:\:Dependent\:\:Events \\[5ex]
P(L'L) = \dfrac{7}{9} * \dfrac{2}{8} = \dfrac{14}{72} ...Multiplication\:\:Rule\:\:for\:\:Dependent\:\:Events \\[5ex]
P(exactly\:\:L) = \dfrac{14}{72} + \dfrac{14}{72} \\[5ex]
P(exactly\:\:L) = \dfrac{14 + 14}{72} \\[5ex]
P(exactly\:\:L) = \dfrac{28}{72} \\[5ex]
P(exactly\:\:L) = 0.388888889 \\[3ex]
P(exactly\:\:L) \approx 0.389 \\[3ex]
\underline{Second\:\:Method} \\[3ex]
P(exactly\:\:L) + P(liked\:\:both) + P(liked\:\:neither) = 1 \\[3ex]
P(exactly\:\:L) + \dfrac{1}{36} + \dfrac{7}{12} = 1 \\[5ex]
P(exactly\:\:L) = 1 - \dfrac{1}{36} - \dfrac{7}{12} \\[5ex]
P(exactly\:\:L) = \dfrac{36}{36} - \dfrac{1}{36} - \dfrac{21}{36} \\[5ex]
P(exactly\:\:L) = \dfrac{36 - 1 - 21}{36} \\[3ex]
P(exactly\:\:L) = \dfrac{14}{36} \\[5ex]
P(exactly\:\:L) = \dfrac{7}{18} \\[5ex]
P(exactly\:\:L) = 0.388888889 \\[3ex]
P(exactly\:\:L) \approx 0.389 \\[3ex]
(e.) \\[3ex]
Two\:\:songs\:\:are\:\:selected \\[3ex]
However,\:\:one\:\:is\:\:replayed \\[3ex]
This\:\:means\:\:one\:\:at\:\:a\:\:time \\[3ex]
This\:\:means\:\:Independent\:\:Events \\[3ex]
This\:\:means\:\:With\:\:Replacement\:\:condition \\[3ex]
P(LL) = \dfrac{2}{9} * \dfrac{2}{9} \\[5ex]
P(LL) = \dfrac{4}{81} \\[5ex]
P(LL) = 0.049382716 \\[3ex]
P(LL) \approx 0.049 \\[3ex]
(f.) \\[3ex]
0.049 \lt 0.05 \\[3ex]
$
Yes, it is unusual that she liked the first two songs that was played in the event that one is replayed before another because the probability that she liked both songs, one after the other, is less than
$5\%$ ... Usual and Unusual Events
$
(g.) \\[3ex]
P(L'L') = \dfrac{7}{9} * \dfrac{7}{9} \\[5ex]
P(L'L') = \dfrac{49}{81} \\[5ex]
P(L'L') = 0.604938272 \\[3ex]
P(L'L') \approx 0.605 \\[3ex]
$
(h.)
We can solve this question in two ways.
Use any method you prefer.
The first method is the use of Multiplication and Addition Rules
Liking exactly $1$ of them means:
She liked the first song AND did not like the second song OR
She did not like the first song AND liked the second song
The second method is to consider all cases/scenarios and sum the probabilities to one
The probability that she liked both songs + the probability that she liked neither song + the probability that she liked exactly one song = $1$
(9.) There is a $27.4\%$ probability that a randomly selected person aged $20$ years or older in the
Community of No Name, Colorado is a jogger.
There is also a $20.2\%$ probability that a randomly selected person aged $20$ years or older is a
female, given that the person jogs.
(a.) Determine the probability that a randomly selected person aged $20$ years or older is a female AND
a jogger.
Round your answer to three decimal places as needed.
(b.) Is it unusual to randomly select a person who is both a female and a jogger?
$
Let: \\[3ex]
Jogger = J \\[3ex]
Female = F \\[3ex]
P(J) = 27.4\% = \dfrac{27.4}{100} = 0.274 \\[5ex]
P(F | J) = 20.2\% = \dfrac{20.2}{100} = 0.202 \\[5ex]
P(F \cap J) = ? \\[3ex]
P(F | J) = \dfrac{P(F \cap J)}{P(J)}...Conditional\:\:Probability \\[5ex]
\dfrac{P(F \cap J)}{P(J)} = P(F | J) \\[5ex]
\rightarrow P(F \cap J) = P(F | J) * P(J) \\[5ex]
P(F \cap J) = 0.202(0.274) \\[3ex]
P(F \cap J) = 0.055348 \\[3ex]
P(F \cap J) \approx 0.055 \\[3ex]
(b.) \\[3ex]
0.055 \gt 0.05 \\[3ex]
$
No, the probability of randomly selecting a person who is both a female and a jogger is NOT unusual because the probability is greater than
$5\%$ ... Usual and Unusual Events
(b.) List all the combinations of the letters $a, b, c, d, e$ taking any two letters at a time.
Verify that the number of combinations in the list corresponds to your answer in (a.)
$
(a.) \\[3ex]
_nC_r = \dfrac{n!}{(n - r)!r!} \\[5ex]
For\:\: _5C_2 \\[3ex]
n = 5 \\[3ex]
r = 2 \\[3ex]
_5C_2 = \dfrac{5!}{(5 - 2)! 2!} \\[5ex]
_5C_2 = \dfrac{5!}{3! 2!} \\[5ex]
_5C_2 = \dfrac{5!}{3! * 2!} \\[5ex]
_5C_2 = \dfrac{5 * 4 * 3!}{3! * 2 * 1} \\[5ex]
_5C_2 = 5 * 2 \\[3ex]
_5C_2 = 10 \\[3ex]
(b.) \\[3ex]
a, b, c, d, e,\:\:taking\:\:two\:\:at\:\:a\:\:time \\[3ex]
5\:\:letters\:\:taking\:\:two\:\:letters\:\:at\:\:a\:\:time \implies _5C_2 \\[3ex]
We\:\:need\:\:to\:\:obtain\:\:10\:\:combinations \\[3ex]
Combinations\:\:are \\[3ex]
ab \\[3ex]
ac \\[3ex]
ad \\[3ex]
ae \\[3ex]
bc \\[3ex]
bd \\[3ex]
be \\[3ex]
cd \\[3ex]
ce \\[3ex]
de \\[3ex]
10\:\:combinations
$
(16.) David intends to create a playlist of six songs.
How many ways can he arrange the six songs in the playlist?
This is a case of the Fundamental Counting Principle
No repetitions case
The first song can take any of the six positions in the list
The second song can take any of the remaining five positions in the list
The third song can take any of the remaining four positions in the list
The fourth song can take any of the remaining three positions in the list
The fifth song can take any of the remaining two positions in the list
The sixth song can take the remaining one position (the last position) in the list
$
Number\:\:of\:\:ways = 6! \\[3ex]
6! = 6 * 5 * 4 * 3 * 2 * 1 \\[3ex]
6! = 720 \\[3ex]
$
David can arrange the six songs in the playlist in $720$ different ways.
(17.) If the correct seven-digit code is entered, a four-key keypad can be used to gain access to a
building.
The digits can be repeated in the code.
(a.) How many codes are possible?
(b.) Determine the probability of entering the correct code on the first try, assuming the owner does
not remember the code?
This is a case of the Fundamental Counting Principle
Numbers can be repeated
(18.) A forty-nine member committee needs to select four members to serve as President, Vice President,
Secretary, and Treasurer respectively.
How many ways can the committee make this selection?
This is a case of Permutation because the selection is with regard to order
The four positions are ordered.
$
_nP_r = \dfrac{n!}{(n - r)!} \\[5ex]
For\:\: _{49}P_4 \\[3ex]
n = 49 \\[3ex]
r = 4 \\[3ex]
_{49}P_4 = \dfrac{49!}{(49 - 4)!} \\[5ex]
_{49}P_4 = \dfrac{49!}{45!} \\[5ex]
_{49}P_4 = \dfrac{49 * 48 * 47 * 46 * 45!}{45!} \\[5ex]
_{49}P_4 = 49 * 48 * 47 * 46 \\[3ex]
_{49}P_4 = 5,085,024 \\[3ex]
$
The committee can fill the positions of a President, Vice President,
Secretary, and Treasurer in any $5,085,024$ ways
(19.) A survey of five hundred randomly selected people found that twenty six people play table tennis.
Determine the probability that a randomly selected person play table tennis.
Type an integer or decimal.
$
Let: \\[3ex]
n(S) = 500 \\[3ex]
Table\:\:Tennis = T \\[3ex]
n(T) = 26 \\[3ex]
P(T) = \dfrac{n(T)}{n(S)} \\[5ex]
P(T) = \dfrac{26}{500} \\[5ex]
P(T) = \dfrac{13}{250} \\[5ex]
P(T) = 0.052 \\[3ex]
$
The probability of randomly selecting a table tennis player is $0.052$
(20.) Mark is preparing for his GRE (Graduate Record Examination) exam.
During testing, he will be allowed to select three schools to which his scores will be sent to no cost.
If there are seven colleges that he is considering, how many different ways can he fill the score report form?
This is a case of Combination because the selection is without regard to order
There is no specific order to list the colleges.
$
_nC_r = \dfrac{n!}{(n - r)!r!} \\[5ex]
For\:\: _7C_3 \\[3ex]
n = 7 \\[3ex]
r = 3 \\[3ex]
_7C_3 = \dfrac{7!}{(7 - 3)! 3!} \\[5ex]
_7C_3 = \dfrac{7!}{4! 3!} \\[5ex]
_7C_3 = \dfrac{7!}{4! * 3!} \\[5ex]
_7C_3 = \dfrac{7 * 6 * 5 * 4!}{4! * 3 * 2 * 1} \\[5ex]
_7C_3 = 7 * 5 \\[3ex]
_7C_3 = 35 \\[3ex]
$
Mark can list the colleges in the score report form in any $35$ ways
