Toss 2 coins. Let X be the number of heads showing. Complete the table below to find the probability distribution for X. $$\begin{array}{c|c} X & P(X)\\\hline 0 & 1/4\\\hline 1 & 1/2\\\hline 2 & 1/4\\ \end{array}$$

Verify that this is a probability distribution

All probabilities $P(X)$ listed are between $0$ and $1$, inclusive, and their sum is one, i.e., $1/4 + 1/2 + 1/4 = 1$Draw a probability histogram. Note that the areas of the rectangles correspond to probability values. Label the histogram with class boundaries and class midpoints.

For each function below, decide whether or not it represents a probability distribution. In the case that any one of these is not a probability distribution, indicate all of the properties that were violated. $$\textrm{a. } \begin{array}{c|c} X & P(X)\\\hline 2 & 0.30\\\hline 4 & -0.50\\\hline 6 & 0.10\\\hline 8 & 1.10\\ \end{array} \hspace{0.8in} \textrm{b. } \begin{array}{c|c} X & P(X)\\\hline 1 & 1/7\\\hline 2 & 2/7\\\hline 3 & 3/7\\\hline 4 & 4/7\\ \end{array} \hspace{0.8in} \textrm{c. } \begin{array}{c|c} X & P(X)\\\hline -1 & 0.30\\\hline 0 & 0.40\\\hline 1 & 0.30\\ \end{array}$$

a) not a probability distribution, there can't be negative probabilites; b) not a probability distribution, the sum of the probabilities ($10/7$) exceeds $1$; c) this is a probability distribution as all probabilities are in $[0,1]$ and they sum $1$.Three coins are tossed. Let $X$ be the number of heads showing. Show a table of values for $P(X)$ and draw the histogram.

$$\begin{array}{c|c} X & P(X)\\\hline 0 & 1/8\\\hline 1 & 3/8\\\hline 2 & 3/8\\\hline 3 & 1/8\\ \end{array}$$Toss three coins. Let $X$ be the number of heads showing. Find the mean and variance.

Using the formulas for any pdf, we find: $$\mu = E(X) = (0)(1/8) + (1)(3/8) + (2)(3/8) + 3(1/8) = 1.5$$ $$\sigma^2 = Var(X) = (0^2)(1/8) + (1^2)(3/8) + (2^2)(3/8) + (3^2)(1/8) - 1.5^2 = 0.75$$ Although things can be found more quickly by realizing we are looking at a binomial distribution here, so $E(X) = np = 3(1/2) = 1.5$ and $Var(X) = npq = 3(1/2)(1/2) = 0.75$Is it unusual to get 3 heads when tossing 3 coins?

The probability is $1/8 = 0.125 \gt 0.05$, so this is not an unusual event.A 35-year-old woman purchases a $\$100,000$ term life insurance policy for an annual payment of $\$360$. Based on a period life table for the U.S. government, the probability that she will survive the year is $0.999057$. Find the expected value of the policy for the insurance company.

If she survives the year, the company makes $\$360$. If not, the company loses $100000 - 360$ dollars, so the distribution looks like: $$\begin{array}{c|c} X & P(X)\\\hline 360 & 0.999057\\\hline -99640 & 0.000943\\ \end{array}$$ Thus, $E(X) = (360)(0.999057) + (-99640)(0.000943) = \$265.7$ is the expected value of the policy.If a player rolls two dice and gets a sum of 2 or 12, she wins $\$20$. If the person gets a $7$, she wins $\$5$. The cost to play the game is $\$3$. Find the expected value of the game.

Keeping in mind that she walks away from game with her winnings minus the cost of the game, and the probabilities of rolling a $2$ or $12$ is $2/36$, rolling a $7$ is $6/36$, and rolling anything else is $1 - 2/36 - 6/36$ (all of which are found upon examination of the sample space for rolling these two dice), we have the following distribution: $$\begin{array}{c|c} X & P(X)\\\hline 17 & 2/36\\\hline 2 & 6/36\\\hline -3 & 28/36\\ \end{array}$$ Thus, $E(X) = (17)(2/36)+(2)(6/36)+(-3)(28/36) \doteq -1.056$.The number of refrigerators sold per day at a local appliance store is shown in the table below, along with the corresponding probabilities. Find the standard deviation. $$\begin{array}{c|c} X & P(X)\\\hline 0 & 0.1\\\hline 1 & 0.2\\\hline 2 & 0.3\\\hline 3 & 0.2\\\hline 4 & 0.2\\ \end{array}$$

First recall, $SD(X) = \sqrt{Var(X)}$.

Now recalling the formula for the variance depends upon the expected value, we find $E(X)$ first. $$E(X) = (0)(0.1)+(1)(0.2)+(2)(0.3)+(3)(0.2)+(4)(0.2) = 2.2$$ so then $$Var(X) = (0^2)(0.1)+(1^2)(0.2)+(2^2)(0.3)+(3^2)(0.2)+(4^2)(0.2) - 2.2^2 = 1.56$$ and finally $$SD(X) = \sqrt{1.56} = 1.249$$

Toss $6$ coins. Let $X$ be the number of heads showing. Find the probability of getting each of the following:

- 4 heads
- at most 2 heads
- at least 5 heads

Binomial. $n=6,p=1/2,q=1/2$.

$\displaystyle{\begin{array}{rcl} a) \quad P(X=4) &=& ({}_6 C_4) (1/2)^4 (1/2)^2\\ &\doteq& 0.2344 \end{array}}$

$\displaystyle{\begin{array}{rcl} b) \quad P(X \le 2) &=& P(0)+P(1)+P(2)\\ &=& ({}_6 C_0) (1/2)^0 (1/2)^6 + ({}_6 C_1) (1/2)^1 (1/2)^5 + ({}_6 C_2) (1/2)^2 (1/2)^4\\ &\doteq& 0.34375 \end{array}}$

$\displaystyle{\begin{array}{rcl} c) \quad P(X \ge 5) &=& P(5) + P(6)\\ &=& ({}_6 C_5) (1/2)^5 (1/2)^1 + ({}_6 C_6) (1/2)^6 (1/2)^0\\ &\doteq& 0.109375 \end{array}}$

Toss 6 coins. Let $X$ be the number of heads. Find the mean and standard deviation of $X$.

Binomial. $n=6,p=1/2,q=1/2$. Thus, the mean is given by $$\mu = E(X) = np = (6)(1/2) = 3$$ and the standard deviation is given by $$\sigma = SD(X) = \sqrt{Var(X)} = \sqrt{npq} = \sqrt{(6)(1/2)(1/2)} \doteq 1.225$$In a survey of 150 senior executives, 48% said that the most common job interview mistake is to have little or no knowledge of the company.

If 3 of those surveyed executives are randomly selected without replacement for a follow-up survey, find the probability that 3 of them said that the most common job interview mistake is to have little or no knowledge of the company.

If 9 of those surveyed executives are randomly selected without replacement for a follow-up survey, explain why the binomial probability formula cannot be used to determine the probability that 4 of them said that the most common job interview mistake is to have little or no knowledge of the company, then find this probability in another way.

a) Since sample size is less than 5% of population sampled (i.e., $3 \lt (150)(0.05) = 7.5$), we can approximate this with the binomial probability formula (i.e., trials are*essentially*independent, even if they are not technically so). Thus, $$P(X=3) = {}_3 C_3 (0.48)^3(0.52)^0 \doteq 0.1106$$ b) Here, the sample size is not less than 5% of the population sampled (again, $7.5$). Thus, we can't use the binomial probability formula to approximate what should really be calculated with combinations. Note there are $(150)(0.48) = 72$ executives that say no knowledge is the mistake, while $(150)(0.52) = 78$ executives that say something else. Thus, if we want the probability that 4 of them say no knowledge is the mistake, then this is given by: $$\frac{ ({}_{72} C_4 )({}_{78} C_5) }{({}_{150} C_9 )} \doteq 0.2618$$A student takes a 10-question multiple-choice exam with four choices for each question and guesses on each question.

Would it be unusual to guess at least 7 out of 10 correctly? (Use the range rule of thumb)

Calculate the probability of guessing at least 7 out of 10 correctly.

a) This is a situation where it would be better to use the binomial probability formula to find the probability exactly (as asked for in part b). That said, if we insist on using the range rule of thumb, which states the standard deviation is approximately one quarter of the range, we have $\sigma = (10 - 0)/4 = 2.5$. Being more than 2 standard deviations from the mean is commonly an unusual event. In this binomial situation, the mean is given by $\mu = E(X) = np = (10)(1/4) = 2.5$. Adding two standard deviations to the mean gives us $2.5 + 2(2.5) = 7.5$, so $7$ does not appear unusual in this circumstance. However, since we are dealing with a binomial distribution, we can quickly calculate the actual standard deviation: $\sigma = \sqrt{npq} = \sqrt{(10)(0.25)(0.75)} \doteq 1.369$. So the mean plus two standard deviations is actually $2.5 + 2(1.369) = 5.238$, suggesting that $7$ really is unusual. All that said, the rule of thumb that we should go out two standard deviations from the mean to find our unusual events actually depends on whether or not the binomial distribution involved can be approximated by a normal distribution -- and that opens up a whole other can of worms. So let's just move on to the next part...

b) As mentioned above, this is a binomial situation. $n=10,p=0.25,q=0.75$. We want $$\begin{array}{rcl} P(X \ge 7) &=& P(7) + P(8) + P(9) + P(10)\\ &=& ({}_{10} C_7) (0.25)^7(0.75)^3 + ({}_{10} C_8) (0.25)^8(0.75)^2 \\ &\quad& + \, ({}_{10} C_9) (0.25)^9(0.75)^1 + ({}_{10} C_{10}) (0.25)^{10}(0.75)^0\\ &\doteq& 0.0035 \end{array}$$ Anything less than $0.05$ is normally considered "unusual", so this event having a probability of $0.0035$ most certainly is!

Suppose you get on average 12 phone calls per day. Assuming phone calls are equally likely to occur at any time of day, find the probability of getting 3 phone calls in one hour.

Poisson. Expected number of calls in one hour is $\lambda = 12/24 = 0.5$, since there are 24 hours in a day. So, $$P(3) = \frac{e^{-0.5} 0.5^3}{3!} \doteq 0.0126$$A truck driver has on average one flat tire every 2000 miles. Find the probability of each of the following:

- exactly 2 flat tires on a 2000 mile trip
- exactly 2 flat tires on a 4000 mile trip
- more than 2 flat tires on a 4000 mile trip
- at most 2 flat tires on a 2000 mile trip

Poisson. Expected number of flat tires in a 2000 mile trip is $\lambda_{2000} = 1$, while the expected number of flat tires in a 4000 mile trip is $\lambda_{4000} = 2$. Thus,a) $\displaystyle{P(2) = \frac{e^{-1} 1^2}{2!} \doteq 0.1839}$

b) $\displaystyle{P(2) = \frac{e^{-2} 2^2}{2!} \doteq 0.2707}$

c)

$\displaystyle{\quad \begin{array}{rcl} P(X > 2) &=& P(3) + P(4) + P(5) + \cdots\\\ &=& 1 - P(0) - P(1) - P(2)\\ &=& \displaystyle{1 - \frac{e^{-2} 2^0}{0!} - \frac{e^{-2} 2^1}{1!} - \frac{e^{-2} 2^2}{2!}} &\doteq& 0.3233 \end{array} }$d)

$\displaystyle{ \quad \begin{array}{rcl} P(X \le 2) &=& P(0) + P(1) + P(2)\\ &=& \displaystyle{\frac{e^{-1} 1^0}{0!} + \frac{e^{-1} 1^1}{1!} + \frac{e^{-1} 1^2}{2!}} &\doteq& 0.1839 \end{array} }$If 3% of all cars fail the emissions inspection, find the probability that in a sample of 150 cars, 4 will fail. (Use the Poisson approximation.)

Note, using the Poisson is appropriate as $n = 150 \ge 100$ and $np = (150)(0.03) = 4.5 \le 10$. The expected number that will fail in $150$ cars is $\lambda = np = 4.5$. So, $$P(4) \approx \frac{e^{-4.5} 4.5^4}{4!} \doteq 0.1898$$There is a 0.9986 probability that a randomly selected 30-year-old mail lives through the year. A life insurance company charges $\$161$ for insuring that the male will live through the year. If the male does not survive the year, the policy pays out $\$100,000$ as a death benefit. Assume that the company sells 1300 such policies so it collects $\$209,300$ in policy payments. The company will make a profit if the number of deaths in this group is 2 or fewer.

What is the mean number of deaths in such groups of 1300 males?

Use the Poisson distribution to find the probability that the company makes a profit from the 1300 policies.

Use the binomial distribution to find the probability that the company makes a profit from the 1300 policies, then compare the result to the result found in part (b).

Binomial. To find mean number of deaths, equate "death" with "success" (yes, that seems a bit morbid). Then $n = 1300, p = 1 - 0.9986 = 0.0014, q = 0.9986$. Then $\mu = np = (1300)(0.0014) = 1.82$

Check to see if approximating the binomial with Poisson is appropriate first. But as $n = 1300 \ge 100$ and $np = 1.82 \le 10$, it is.

Thus, $\lambda = 1.82$ and $$P(\textrm{profit}) = P(2 \textrm{ or fewer die}) = P(X \le 2) = P(0) + P(1) + P(2)$$ Finding this sum, we have $$\begin{array}{rcl} P(0)+P(1)+P(2) &=& \displaystyle{\frac{e^{-1.82} 1.82^0}{0!} + \frac{e^{-1.82} 1.82^1}{1!} + \frac{e^{-1.82} 1.82^2}{2!}}\\\\ &=& \displaystyle{e^{-1.82} \left(1 + 1.82 + \frac{1.82^2}{2}\right)}\\\\ &\doteq& 0.72525 \end{array}$$

Worked directly as a binomial, we have $$\begin{array}{rcl} P(0)+P(1)+P(2) &=& ({}_1300 C_0)(0.0014)^{0}(0.9986)^{1300}\\ &\quad& + \, ({}_1300 C_1)(0.0014)^{1}(0.9986)^{1299}\\ &\quad& + \, ({}_1300 C_2)(0.0014)^{2}(0.9986)^{1298}\\ &\doteq& 0.72529 \end{array}$$

A shipment of 24 computer keyboards is rejected if 4 are checked for defects and at least 1 is found to be defective. Find the probability that the shipment will be returned if there are actually 6 defective keyboards.

Hypergeometric. There are 6 defective keyboards, so there are 18 good keyboards. Thus $$P(X \ge 1 \textrm{ defective}) = 1 - P(0 \textrm{ defective}) = 1 - \frac{({}_{18} C_4)({}_6 C_0)}{{}_{24} C_4} \doteq 0.7120$$Three cards are drawn without replacement from a standard deck of 52 cards. Find the probability of getting

- all hearts
- at least 2 hearts
- all hearts, given that all three cards are red

$\displaystyle{P(3) = \frac{({}_{13} C_3)({}_{39} C_0)}{{}_{52} C_3} \doteq 0.0129}$

$\displaystyle{P(2)+P(3) = \frac{({}_{13} C_2)({}_{39} C_1)}{{}_{52} C_3} + \frac{({}_{13} C_3)({}_{39} C_0)}{{}_{52} C_3} \doteq 0.1506}$

$\displaystyle{P(3) = \frac{({}_{13} C_3)({}_{13} C_0)}{{}_{26} C_3} = 0.11}$

Three cards are drawn with replacement from a standard deck of 52 cards. Find the probability of getting

- all hearts
- at least 2 hearts
- all hearts, given that all three cards are red

Parts (a) and (b): Binomial with $n=3,p=1/4,q=3/4$$P(3) = ({}_3 C_3)(1/4)^3(3/4)^0 = 0.015625$

$P(X \ge 2) = P(2) + P(3) = ({}_3 C_2) (1/4)^2 (3/4)^1 + ({}_3 C_3)(1/4)^3(3/4)^0 = 0.15625$

Binomial. $n=3,p=1/2,q=1/2$, so $P(3) = ({}_3 C_3)(1/2)^3(1/2)^0 = 0.125$

A book of 1000 pages contains 200 typos. What is the probability that page 13 contains exactly 3 typos?

This is technically binomial, but approximating it as a Poisson will make calculations a bit easier. To see how it is binomial, consider the task of "sprinkling" in the 200 typos amongst the 1000 pages. The probability that any one of these 200 typos lands on page $13$ is $1/1000$. Calling a typo on page $13$ a "success", we are interested in the probability of $3$ successes in $n = 200$ trials, each with probability $p = 1/1000$ (and then $q = 999/1000$). However, $n = 200 \ge 100$ and $np = 0.2 \le 10$, so approximation with Poisson is appropriate. Thus, using $\lambda = 0.2$ we have $$P(3) = \frac{e^{-0.2} 0.2^3}{3!} \doteq 0.0011$$