1)

A box contains 54 red marbles, 9 white marbles, and 75 blue marbles. If a marble is randomly selected from the box, what is the probability that it is red or blue? Express your answer as a fraction or a decimal number rounded to four decimal places.

JB First I find the probability that the pulled marble is red, \(P(R)=\frac{54}{138}=\frac{9}{23}\). Then I find the probability that the pulled marble is blue \(P(B)=\frac{75}{138}=\frac{24}{46}\). Then, to find the probability that a pull is red or blue, I use: \(P(R \cup B)= P(R)+P(B)=\frac{9}{23}+\frac{24}{46}=\frac{43}{46}\)

2)

You are going to play mini golf. A ball machine that contains 19 green golf balls, 20 red golf balls, 24 blue golf balls, and 17 yellow golf balls, randomly gives you your ball. What is the probability that you end up with a red golf ball?

JB \(P(R)=\frac{20}{19+20+24+17}=\frac{20}{80}=\frac{1}{4}\)

3)

JB First, I find the probability that a customer is not male which is the same as the probability that the customer is female: \(P(F)=\frac{female}{total}=\frac{728}{1399}\). Next I find the probability that a customer does not live with parents: \(P(NoParents)=\frac{NoParents}{Total}=\frac{apt+ dorm+frat+other}{total}=\frac{81+116+130+129}{1399}=\frac{456}{1399}\). Then the probability that a customer is not male and does not live with their parents is:

\(P(female \cup NoParents)= P(female)+P(NoParents)=\frac{728}{1399}+\frac{456}{1399}=\frac{1184}{1399}=0.8463\)

4)

Determine if going to the gym and losing weight are independent:

JB A: Dependent.

The events A and B are independent if B has no effect on A or A has no effect on B.

5)

Wrap contains 3 diff veggies, 3 diff condiments, in a tortilla. If there are 3 diff tortillas, 8 diff veggies, and 7 types of condiments, how many diff wraps can be made?

JB By the multiplication rule, if one operation can be done in \(n_1\) different ways and a second operation can be done in \(n_2\) different ways, then the number of different ways they can both be done is \(n_1 \times n_2\). This leads me to: \(3 \times 8 \times 7 =168\) different wraps that can be made.

6)

Determine if the following are independent: Jeff runs out of gas on way to work and Liz watches the evening news.

JB B. They’re independent.

The events A and B are independent if B has no effect on A or A has no effect on B.

7)

The newly elected president needs to decide the remaining 8 spots available in the cabinet he/she is appointing. If there are 14 eligible candidates for these positions (where rank matters), how many different ways can the members of the cabinet be appointed?

JB The question is asking how many different ways can the 8 remaining spots be filled by the 14 eligible where order (rank) matters. In this case, I’ll use k-permutations which can tell me how many different \(k\) groupings can be selected from \(n\) elements where order matters is defined as \(\frac{n!}{(n-k)!}=\frac{14!}{(14-8)!}=121,080,960\) different ways.

8)

A bag contains 9 red, 4 orange, and 9 green jellybeans. What is the probability of reaching into the bag and randomly withdrawing 4 jellybeans such that the number of red ones is 0, the number of orange ones is 1, and the number of green ones is 3? Write your answer as a fraction or a decimal number rounded to four decimal places.

JB First I get the total number of events possible i.e. the number of ways that I can pull out 4 different beans from the bag as: \(\binom{22}{4}=\frac{22!}{4!(22-4)!}=7,315\). Within a 4-bean pull, the different ways that I can get 0 red, 1 orange, and 3 green are: \(\binom{9}{0}_{red} \times \binom{4}{1}_{orng} \times \binom{9}{3}_{grn} =1\times4 \times84=336\). This leads me to the probability \(P(noRed,1Orange,3Green)=\frac{336}{7315}=\frac{48}{1045} \approx 0.0459\)

9)

Evaluate the following expression: \(\frac{11!}{7!}\):

JB \(\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times \times 4 \times 3 \times 2 \times 1 }{7 \times 6 \times 5 \times \times 4 \times 3 \times 2 \times 1 }=11 \times 10 \times 9 \times 8=7,920\):

10)

Describe the compliment of the event: 67% of subscribers to a fitness magazine are over the age of 34.

JB “33% of subscribers to a fitness magazine are under the age of 34.”

11)

If you throw exactly three heads in four tosses of a coin you win $97. If not, you pay me $30.

Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
Step 2. If you played this game 559 times how much would you expect to win or lose? (Losses must be entered as negative.)

JB S1 A good reference is this Khan Academy video. First, we find out the total number of outcomes from 4 tosses of a coin which is: Total Qty Outcomes \(=(Qty/Toss)^n=2^4=16\). Out of 4 of these flips, we’re choosing 3 of them to be heads so to find that we calculate \(\binom{4}{3}=\frac{4!}{3!(4-3)!}=4\) different ways to get an outcome of 3 heads. Then the probability of getting 3 heads in 4 tosses is: \(\frac{4}{16}=\frac{1}{4}\). Then the expected value of the proposition would be: \(P(Win)\times (\$97) + P(Lose)\times (-\$30)=\frac{1}{4}\times (\$97)+(1-\frac{1}{4})\times (-\$30)=\$1.75\) winnings.

JB S2 The expected $-amount to win after 559 games would be: \(\$1.75 \times 559=\$978.25\) which seems like a decent income for coin-flipping.

12)

Flip a coin 9 times. If you get 4 tails or less, I will pay you $23. Otherwise you pay me $26.

Step 1. Find the expected value of the proposition. Round your answer to two decimal places. Step 2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)

JB S1 First I get the total number of possible outcomes among 9 tosses: \(2^n=2^9=512\). Then, out of 9 tosses, I “win” if I get fewer than or equal to 4 tails which could be expressed like \(P(4,3,2,1, or 0 T) = P(4T)+P(3T)+P(2T)+P(1T)+P(0T)=\)

\(\frac{\binom{9}{4}}{512}+\frac{\binom{9}{3}}{512}+\frac{\binom{9}{2}}{512}+\frac{\binom{9}{1}}{512}+\frac{\binom{9}{0}}{512}\). This leads to the calculation:

\(\frac{126}{512}+\frac{84}{512}+\frac{36}{512}+\frac{9}{512}+\frac{1}{512}=\frac{1}{2}\) which is the probability of getting 4 or less among 9 tosses which yeilds and expected value of the proposition as:

\(P(x\leq4T_{9tosses}) \times \$win+(1-P(x\leq4T_{9tosses}))(-\$lose)=\)

\(\frac{1}{2} \times \$23 + \frac{1}{2} \times-\$26=-\$\frac{3}{2}\) which indicates the expected values is a loss of $1.50.

JB S2 After 994 games, I’d expect to lose: \(-\$1.50 \times994=-\$1,491\) which is not a good way to go.

13)

The sensitivity and specificity of the polygraph has been a subject of study and debate for years. A 2001 study of the use of polygraph for screening purposes suggested that the probability of detecting a liar was .59 (sensitivity) and that the probability of detecting a “truth teller” was .90 (specificity). We estimate that about 20% of individuals selected for the screening polygraph will lie.

1. What is the probability that an individual is actually a liar given that the polygraph detected him/her as such? (Show me the table or the formulaic solution or both.)

2. What is the probability that an individual is actually a truth-teller given that the polygraph detected him/her as such? (Show me the table or the formulaic solution or both.)

3. What is the probability that a randomly selected individual is either a liar or was identified as a liar by the polygraph? Be sure to write the probability statement.

JB a \(P(Liar|Poly_{liar})=\frac{P(Poly_{liar}|Liar)P(Liar)}{P(Poly_{liar})}=\frac{P(Poly_{liar}|Liar)P(Liar)}{P(Poly_{liar}|Liar)P(Liar)+P(Poly_{liar}|Truth)P(Truth)}\)

\(=\frac{.59\times .20}{.59\times .20 + 0.1 \times.8}\approx 0.59596\)

JB b \(P(Truth|Poly_{truth})=\frac{P(Poly_{truth}|Truth)}{P(Poly_{truth}|Truth) + P(Poly_{truth}|Liar)P(Liar)}=\frac{.9 \times .8}{.9 \times .8 + .41\times.2} \approx 0.8978\)

JB c \(P(Liar \cup Poly_{liar}))=P(Liar) + P(Poly_{liar})=.2+(.8\times.1+.2\times(.41)) =.3620\)