Company A supplies 40% of the computers sold and is late 5% of the time. Company B supplies 30% of the computers sold and is late 3% of the time. Company C supplies another 30% and is late 2.5% of the time. A computer arrives late - what is the probability that it came from Company A?

Bayes Theorem. Make a tree: $P(L) = 0.0365$ and $P(A \textrm{ and } L) = (0.4)(0.05) = 0.02$, so P(shipped from A given that the computer is late) = 0.548, approximately.In Orange County, 51% of the adults are males. One adult is randomly selected for a survey involving credit card usage. It is later learned that the selected survey subject was smoking a cigar. Also, 9.5% of males smoke cigars, whereas 1.7% of females smoke cigars (based on data from the Substance Abuse and Mental Health Services Administration). Use this additional information to find the probability that the selected subject is a male.

$$P(M|S) = \frac{(0.51)(0.095)}{(0.51)(0.095)+(0.49)(0.017)} \doteq 0.853$$A person uses his car 30% of the time, walks 30% of the time and rides the bus 40% of the time as he goes to work. He is late 10% of the time when he walks; he is late 3% of the time when he drives; and he is late 7% of the time he takes the bus.

- What is the probability he took the bus if he was late?
- What is the probability he walked if he is on time?

- $\displaystyle{P(B|L) = \frac{(0.40)(0.07)}{(0.40)(0.07)+(0.30)(0.03)+(0.30)(0.10)} \doteq 0.418}$
- $\displaystyle{P(W|T) = \frac{(0.30)(.90)}{(0.30)(0.97)+(0.30)(0.90)+(0.40)(0.93)} \doteq 0.289}$

In a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.

- If one of the study subjects is randomly selected, find the probability of getting someone who was not sent to prison.
- If a study subject is randomly selected and it is then found that the subject entered a guilty plea, find the probability that this person was not sent to prison.

$1-0.45 = 0.55$

$\displaystyle{\frac{(0.55)(0.55)}{(0.45)((0.40)+(0.55)(0.55)} \doteq 0.627}$

On a game show, a contestant can select one of four boxes. The red box contains one $\$100$ bill and nine $\$1$ bills. A green box contains two $\$100$ bills and eight $\$1$ bills. A blue box contains three $\$100$ bills and seven $\$1$ bills. A yellow box contains five $\$100$ bills and five $\$1$ bills. The contestant selects a box at random and selects a bill from the box at random. If a $\$100$ bill is selected, find the probability that it came from the yellow box.

$\displaystyle{\frac{(0.25)(0.50)}{(0.25)(0.10)+(0.25)(0.20)+(0.25)(0.30)+(0.25)(0.50)} \doteq 0.455}$A plane's "black-box" is manufactured by only 3 companies: AirCorp, BigSkies, and CharterUS - who make 80%, 15%, and 5% of all the black-boxes made, respectively. Invariably, some of these are defective. Assuming the percentage of defective black-boxes made by AirCorp, BigSkies, and CharterUS are 4%, 6%, and 9%, repsectively, find the probability that a randomly selected black-box from all black-boxes made that is found to be defective came from AirCorp.

$\displaystyle{\frac{(0.80)(0.04)}{(0.80)(0.04)+(0.15)(0.06)+(0.05)(0.09)} \doteq 0.7033}$Consider 3 coins where two are fair, yielding heads with probability $0.50$, while the third yields heads with probability $0.75$. If one randomly selects one of the coins and tosses it 3 times, yielding 3 heads - what is the probability this is the biased coin?

First note two things: 1) the probability of drawing a fair coin is $2/3$ and the probability of drawing a biased coin is $1/3$; and 2) the probability of tossing 3 heads with a fair coin is $(1/2)^3 = 0.125$, while the probability of tossing 3 heads with the described biased coin is $(0.75)^3$. Then, use Baye's Theorem: $$\displaystyle{\frac{(1/3)(0.75)^3}{(2/3)(1/2)^3+(1/3)(0.75)^3} \doteq 0.6279}$$Suppose $P(A), P(\overline{A}), P(B|A)$, and $P(B|\overline{A})$ are known. Find an expression for $P(A|B)$ in terms of these four probabilities.

$\displaystyle{\frac{P(A)P(B|A)}{P(A)P(B|A) + P(\overline{A})P(B|\overline{A})}}$Assume the probability of having tuberculosis (TB) is 0.0005, and a test for TB is 99% accurate. What is the probability one has TB if one tests positive for the disease?

$\displaystyle{\frac{(0.0005)(0.99)}{(0.0005)(0.99)+(0.9995)(0.01)} \doteq 0.0472}$An automobile manufacturer has three factories: A, B, and C. They produce 50%, 30%, and 20% respectively, of a specific model of car. 30% of the cars produced in factory A are white, 40% of those produced in factory B are white, and 25% produced in factory C are white.

- If an automobile produced by the company is selected at random, find the probability that it is white.
- Given that an automobile selected at random is white, find the probability that it came from factory B.

$(0.50)(0.30)+(0.30)(0.40)+(0.20)(0.25) = 0.32$

Given the calculation in part (a), we have $\displaystyle{\frac{(0.30)(0.40)}{0.32} = 0.375}$

Two manufacturers supply blankets to emergency relief organizations. Manufacturer A supplies 3000 blankets and 4% are irregular in workmanship. Manufacturer B supplies 2400 blankets and 7% are found to be irregular. Given that a blanket is irregular, find the probability that it came from manufacturer B.

$\displaystyle{\frac{\left(\frac{2400}{5400}\right)(0.07)}{\left(\frac{2400}{5400}\right)(0.07)+\left(\frac{3000}{5400}\right)(0.04)} \doteq 0.5833}$