Exercises - Probability Distributions

Determine which of the following represent valid probability mass functions.

$\displaystyle{\begin{array}{c|c|c|c|c}x & 0 & 1 & 2 & 3 \\\hline P(x) & 1/8 & 3/8 & 3/8 & 1/8\end{array}}$
$\displaystyle{\begin{array}{c|c|c|c|c}x & 0 & 1 & 2 & 3 \\\hline P(x) & 1/8 & -3/8 & 3/4 & 1/2\end{array}}$
$\displaystyle{\begin{array}{c|c|c|c|c}x & 0 & 1 & 2 & 3 \\\hline P(x) & 1/8 & 3/8 & 3/8 & 2/8\end{array}}$

(a) this IS a valid probability mass function as the probabilities listed are always between 0 and 1, inclusive, and the probabilities sum to 1; (b) NOT a valid probability mass function, as $P(1)$ is not between 0 and 1, inclusive; (c) NOT a valid probability mass function as the sum of the probabilities is not equal to 1.

Is the function described below a valid probability mass function? Explain.

$$f(x)=\frac{x+1}{10} \textrm{ for } x = 0, 1, 2, 3$$

Note $0 \le f(x) \le 1$ for $x = 0, 1, 2, 3$ and $f(0) + f(1) + f(2) + f(3) = 1$, so $f(x)$ does indeed describe a probability mass function.

Find the mean, variance, and standard deviation for the following distribution

$$\displaystyle{\begin{array}{c|c|c|c|c}x & 0 & 1 & 2 & 3 \\\hline P(x) & 1/8 & 3/8 & 3/8 & 1/8\end{array}}$$

mean: $\mu = (0)(1/8) + (1)(3/8) + (2)(3/8) + (3)(1/8) = 12/8 = 3/2$;
variance: $\sigma^2 = (0^2)(1/8) + (1^2)(3/8) + (2^2)(3/8) + (3^2)(1/8) - \mu^2 = 3/4$;
standard deviation: $\sigma = \sqrt{\sigma^2} \doteq 0.866$

A game at a local fair will give $\$$100 to anyone that can break a balloon by throwing a dart at it. The game costs $\$$5 to throw a single dart, and you're willing to spend $\$$30 trying to win. Assuming that you have a 10% chance of hitting the balloon on any given throw, find the expected number of darts you will throw.

Consider the possible outcomes, we could throw a dart 1, 2, 3, 4, 5, or 6 times. Letting these be the possible values for $x$, we find $P(x)$, the probability that we throw exactly $x$ darts for each such value.

We throw exactly 1 dart precisely when we break the balloon on our first attempt. Thus, $P(1) = 0.10$. We throw exactly 2 darts when we fail to break the balloon on the first attempt, but succeed on the second. So $P(2) = (0.90)(0.10)$, recalling that if we have a 10% chance of hitting the balloon on any given throw, we have a 90% chance of missing it. Similarly, we find $P(3) = (0.90)^2(0.10)$, $P(4) = (0.90)^3(0.10)$, and $P(5) = (0.90)^4(0.10)$.

However, we find the probability of throwing 6 darts a little differently, as one throws exactly 6 darts when one misses the previous 5. It doesn't matter whether we hit the balloon on the sixth try or not, as we are only willing to spend $\$30$ trying to win, and will stop throwing darts after the sixth try regardless. Consequently $P(6) = (0.90)^5$.

Collecting these results in a table, we have:

$$\begin{array}{c|c|c|c|c|c|c} x & 1 & 2 & 3 & 4 & 5 & 6\\\hline P(x) & 0.10 & (0.90)(0.10) & (0.90)^2(0.10) & (0.90)^3(0.10) & (0.90)^4(0.10) & (0.90)^5 \end{array}$$

Now, calculating the expected value $\mu$, we have

$$\begin{array}{rcl} \mu &=& (1)(0.10) + (2)(0.90)(0.10) + (3)(0.90)^2(0.10)\\ & & + \, (4)(0.90)^3(0.10) + (5)(0.90)^4(0.10) + (6)(0.90)^5\\ &=& 4.68559 \end{array}$$

Based on previous information, it has been determined that 90% of the population brush their teeth once a day. Answer the following for a sample of 20 people:

What is the probability that exactly 18 people brush their teeth once a day?
What is the probability that at least 18 people brush their teeth once a day?

Treat as a binomial distribution, with $n = 20, p = 0.90,$ and $q = 0.10$:

(a) $$P(18) = {}_{20}C_{18} (0.90)^{18}(0.10)^2$$

In R:
dbinom(18,size=20,prob=0.90)
[1] 0.2851798

In Excel:
=BINOM.DIST(18,20,0.9,FALSE)

(b) $$\begin{array}{rcl} P(18) + P(19) + P(20) &=& {}_{20}C_{18} (0.90)^{18}(0.10)^2\\ && + \, {}_{20}C_{19} (0.90)^{19}(0.10)^1\\ && + \, {}_{20}C_{20} (0.90)^{20}(0.10)^0 &\approx& 0.6769 \end{array}$$

In R:
1-pbinom(17,size=20,prob=0.90)
[1] 0.6769268

In Excel:
=1-BINOM.DIST(17,20,0.9,TRUE)

A fair coin is tossed eight times. Let the random variable be the number of heads that appear.

What is the probability that exactly 4 heads appear?
What is the probability that you get either all heads or all tails?

This is binomial with $n = 8$, and $p = 1/2$. (a) $70/256 \approx 0.2734$; (b) $1/256 + 1/256 = 1/128 \approx 0.0078$

Toss a coin 16 times. Let X be the number of heads that appear.

Find the probability that there will be more than 13 heads using a Binomial distribution.

This is binomial with $n = 16$ and $p = 1/2$. (a) $P(14) + P(15) + P(16) \approx 0.0021$;

Roll a standard die 8 times. Let $X$ be the number of 2's rolled. Find the following:

$P(X=3)$, the probability of rolling exactly 3 two's
$P(X \lt 3)$, the probability of rolling less than 3 two's
$P(X \ge 3)$, the probability of rolling at least 3 two's

(a) $P(3) \approx 0.1042$; (b) $P(0) + P(1) + P(2) \approx 0.8652$; (c) $1 - P(X \lt 3) \approx 0.1348$

A shipment of 25 computers contains 10 computers with a defective DVD burner. What is the probability, if a random sample of 6 computers is selected and then tested, that the sample will contain at least 1 defective computer?

$$P( \textrm{ at least 1 defective} ) = 1 - P(0 \textrm{ defective }) = 1 - \frac{({}_{15}C_6)({}_{10}C_0)}{({}_{25}C_6)} \approx 0.9717$$

In a club of 25 members there are 10 married men and 15 unmarried men. What is the probability that a subcommittee of 6 will have at least 1 married man?

$$P( \textrm{ at least 1 married man} ) = 1 - P(0 \textrm{ married men }) = 1 - \frac{({}_{10}C_0)({}_{15}C_6)}{({}_{25}C_6)} \approx 0.9717$$

It is known that 5% of all tax returns contain at least one error. For a random selection of 10 tax returns, what is the probability that at most 2 of them contain errors?

This is a binomial with $n = 10$ and $p = 0.05$. $P(0) + P(1) + P(2) = 0.9885$ approximately.

In R:
pbinom(2,size=10,prob=0.05)
[1] 0.9884964

In Excel:
=BINOM.DIST(2,10,0.05,TRUE)

Assume 75% of Americans wear seatbelts. If 200 Americans are selected at random, find the expected number of people in this group that wear their seatbelts.

This is binomial with $n = 200$ and $p = 0.75$. Recall for a binomial distribution, $\mu = np$, so the expected number of people in this group that wear their seatbelts is $(200)(0.75) = 150$.

Toss a fair coin 100 times. Let X be the number of heads showing. Give the mean and the standard deviation for this experiment.

This is binomial with $n = 100$ and $p = 0.50$. $\mu = np = 50$ and $\sigma = \sqrt{npq} = \sqrt{(100)(0.50)(0.50)} = 5$.

The probabilities that a game of chance results in a win, loss, or tie for the player to go first is 0.48, 0.46, and 0.06, respectively. If the game is played 8 times, find the probability that there will be 3 wins, 4 losses and 1 tie.

Multinomial distribution. $$P( \textrm{3 wins, 4 losses, 1 tie} ) = \frac{8!}{3! 4! 1!} (0.48)^3 (0.46)^4 (0.06)^1 \approx 0.0831$$

A DVD has a defect on average every 2 inches along its track. What is the probability of seeing less than 3 defects within a 5 inch section of its track?

Poisson distribution with $\lambda = 5/2$. $$\begin{array}{rcl} P(x \lt 3) &=& P(0) + P(1) + P(2)\\ &=& \displaystyle{\frac{e^{-5/2} (5/2)^0}{0!} + \frac{e^{-5/2} (5/2)^1}{1!} + \frac{e^{-5/2} (5/2)^2}{2!}}\\ &=& \displaystyle{e^{-5/2}\left(1 + \frac{5}{2} + \frac{25}{8}\right)}\\ &\approx& 0.5438 \end{array}$$

In R:
ppois(2,lambda=5/2)
[1] 0.5438131

In Excel:
=POISSON.DIST(2,5/2,TRUE)

Usually 50 potential jurors are held to compose a jury of 12. Suppose that this group of 50 has 15 females and 35 males.

What is the probability that the jury will be made up of all the same sex?
What is the probability that the jury will be made up of 4 females and 8 males?

Hypergeometric. (a) $P(\textrm{all females}) + P(\textrm{all males}) \approx 0.00687$; (b) $\displaystyle{\frac{({}_{15}C_4)({}_{35}C_8)}{({}_{50}C_{12})} \approx 0.2646}$

Compare and contrast the Poisson distribution with the Binomial distribution.

Check notes online.

The probability that a worker will become disabled in a one-year period is 0.0045. If there are 500 workers on an assembly line, find the probability that more than 4 workers will become disabled. (Use the Poisson distribution to approximate the probability.)

Poisson distribution. $\lambda \approx (500)(0.0045) = 2.25$. Use complement: $P(\textrm{more than 4}) = 1 - P(4 \textrm{ or less}) = 1 - (P(0)+P(1)+P(2)+P(3)+P(4))$, which is approximately $1 - 0.92198 \approx 0.078$

Assume that customers enter a large store at the rate of 60 per hour (one per minute). Find the probability that during a five-minute interval no one will enter the store.

Poisson distribution. $\lambda = 5$ (for the five minute period). $P(0) \approx 0.0067$ that no one arrives in the five minute period.

Given the function defined below,

$$f(x) = \frac{x+1}{6} \textrm{ for } x=0,1,2$$

Explain clearly why this is a valid probability mass function.
Find the mean and standard deviation of this function.

(a) Both conditions are met: (i) The sum of the probabilities equals one: $1/6 + 2/6 + 3/6 = 1$, and (ii) $0 \le f(x) \le 1$ for each possible value of $x$; (b) $\mu = 8/6, \sigma = \sqrt{5/9} \approx 0.745$.

You have a box with 25 different colored balls as follows: 4 red, 6 blue, 10 white, and 5 green.

What is the probability that there will be exactly one of each color if four balls are drawn?
What is the probability that there will be two white balls if two balls are drawn?

Hypergeometric.

$\displaystyle{\textrm{(a) } \frac{({}_{4}C_1)({}_{6}C_1)({}_{10}C_1)({}_{5}C_1)}{{}_{25}C_4} = \frac{1200}{12650} \approx 0.095}$

$\displaystyle{\textrm{(b) } \frac{({}_{10}C_2)({}_{15}C_0)}{{}_{25}C_2} = \frac{45}{300} = 0.15}$

Assume that the probability of a college student having a car on campus is 0.30. A random sample of 12 students is taken. What is the probability that at least 4 will have a car on campus?

Work the problem using the Binomial distribution.

(a) Working as a binomial, we use the complement. $1 - (P(0)+P(1)+P(2)+P(3)) \approx 1 - 0.49251 \approx 0.51$;

A bridge hand contains 13 cards. What is the probability that a bridge hand will contain 9 spades, all four aces, and one non-spade, non-ace.

Hypergeometric. $2.8 \times 10^{-8}$

The switchboard receives an average of 3 calls per minute. For a randomly selected minute, what is the probability that there will be at least 4 calls?

Poisson. $1 - (P(0)+P(1)+P(2)+P(3)) \approx 1 - 0.0647 = 0.353$