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Toss 2 coins. Let X be the number of heads showing. Complete the table below to find the probability mass function 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 legitimate probability mass function
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. $$\textrm{a. } \begin{array}{c|c} X & P(X)\\\hline 2 & 0.30\\\hline 4 & -0.50\\\hline 6 & 0.10\\\hline 8 & 1.10\\ \end{array} \hspace{0.8in} \textrm{b. } \begin{array}{c|c} X & P(X)\\\hline 1 & 1/7\\\hline 2 & 2/7\\\hline 3 & 3/7\\\hline 4 & 4/7\\ \end{array} \hspace{0.8in} \textrm{c. } \begin{array}{c|c} X & P(X)\\\hline -1 & 0.30\\\hline 0 & 0.40\\\hline 1 & 0.30\\ \end{array}$$
Three coins are tossed. Let $X$ be the number of heads showing. Show a table of values for $P(X)$ and draw the histogram.
Toss three coins. Let $X$ be the number of heads showing. Find the mean and variance.
Is it unusual to get 3 heads when tossing 3 coins?
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 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.
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:
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 $X$.
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 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 $$\begin{array}{rcl} P(X \ge 7) &=& P(7) + P(8) + P(9) + P(10)\\ &=& ({}_{10} C_7) (0.25)^7(0.75)^3 + ({}_{10} C_8) (0.25)^8(0.75)^2 \\ &\quad& + \, ({}_{10} C_9) (0.25)^9(0.75)^1 + ({}_{10} C_{10}) (0.25)^{10}(0.75)^0\\ &\doteq& 0.0035 \end{array}$$
In R: 1-pbinom(6,size=10,prob=0.25) [1] [1] 0.003505707 In Excel: 1-BINOM.DIST(6,10,0.25,TRUE)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.
In R: dpois(3,0.5) [1] 0.01263606 In Excel: =POISSON.DIST(3,0.5,FALSE)
A truck driver has on average one flat tire every 2000 miles. Find the probability of each of the following:
a) $\displaystyle{P(2) = \frac{e^{-1} 1^2}{2!} \doteq 0.1839}$
In R: dpois(2,1) [1] 0.1839397 In Excel: =POISSON.DIST(2,1,FALSE)
b) $\displaystyle{P(2) = \frac{e^{-2} 2^2}{2!} \doteq 0.2707}$
c)
$\displaystyle{\quad \begin{array}{rcl}
P(X > 2) &=& P(3) + P(4) + P(5) + \cdots\\\
&=& 1 - P(0) - P(1) - P(2)\\
&=& \displaystyle{1 - \frac{e^{-2} 2^0}{0!} - \frac{e^{-2} 2^1}{1!} - \frac{e^{-2} 2^2}{2!}}
&\doteq& 0.3233
\end{array} }$
In R: 1-ppois(2,2) [1] 0.3233236 In Excel: =1-POISSON.DIST(2,2,TRUE)
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.)
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 $$\begin{array}{rcl} P(0)+P(1)+P(2) &=& \displaystyle{\frac{e^{-1.82} 1.82^0}{0!} + \frac{e^{-1.82} 1.82^1}{1!} + \frac{e^{-1.82} 1.82^2}{2!}}\\\\ &=& \displaystyle{e^{-1.82} \left(1 + 1.82 + \frac{1.82^2}{2}\right)}\\\\ &\doteq& 0.72525 \end{array}$$
In R: ppois(2,lambda=1.82) [1] 0.7252597 In Excel: =POISSON.DIST(2,1.82,TRUE)
Worked directly as a binomial, we have $$\begin{array}{rcl} P(0)+P(1)+P(2) &=& ({}_{1300} C_0)(0.0014)^{0}(0.9986)^{1300}\\ &\quad& + \, ({}_{1300} C_1)(0.0014)^{1}(0.9986)^{1299}\\ &\quad& + \, ({}_{1300} C_2)(0.0014)^{2}(0.9986)^{1298}\\ &\doteq& 0.72529 \end{array}$$
In R: pbinom(2,size=1300,prob=0.0014) [1] 0.7252936 In Excel: =BINOM.DIST(2,1300,0.0014,TRUE)
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.
In R: 1-dhyper(0,6,18,4) [1] 0.7120271 In Excel: =1-HYPGEOM.DIST(0,4,6,24,FALSE)
Three cards are drawn without replacement from a standard deck of 52 cards. Find the probability of getting
$\displaystyle{P(3) = \frac{({}_{13} C_3)({}_{39} C_0)}{{}_{52} C_3} \doteq 0.0129}$
$\displaystyle{P(2)+P(3) = \frac{({}_{13} C_2)({}_{39} C_1)}{{}_{52} C_3} + \frac{({}_{13} C_3)({}_{39} C_0)}{{}_{52} C_3} \doteq 0.1506}$
$\displaystyle{P(3) = \frac{({}_{13} C_3)({}_{13} C_0)}{{}_{26} C_3} = 0.11}$
Three cards are drawn with replacement from a standard deck of 52 cards. Find the probability of getting
$P(3) = ({}_3 C_3)(1/4)^3(3/4)^0 = 0.015625$
$P(X \ge 2) = P(2) + P(3) = ({}_3 C_2) (1/4)^2 (3/4)^1 + ({}_3 C_3)(1/4)^3(3/4)^0 = 0.15625$
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?