# Probability Distributions

- Page ID
- 218

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)## Variables

A variable whose value depends upon a chance experiment is called a random variable. Suppose that a person is asked who that person is closest to: their mother or their father. The random variable of this experiment is the boolean variable whose possibilities are {Mother, Father}. A continuous random variable is a variable whose possible outcomes are part of a continuous data set.

The random variable that represents the height of the next person who walks in the room is a continuous random variable while the random variable that represents the number rolled on a six sided die is not a continuous random variable. A random variable that is not continuous is called a *discreet random variable*.

## Probability Distributions

Suppose we toss two dice. We will make a table of the probabilities for the sum of the dice. The possibilities are:

2,3,4,5,6,7,8,9,10,11,12.

\(x\) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
---|---|---|---|---|---|---|---|---|---|---|---|

\(P(x)\) | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |

Suppose that you buy a raffle ticket for $5. If 1,000 tickets are sold and there are 10 third place winners of $25, three second place winners of $100 and 1 grand prize winner of $2,000, construct a probability distribution table. Do not forget that if you have the $25 ticket, you will have won $20.

## Value (Mean)

We when we buy insurance in black jack we lose the insurance bet if the dealer does not have black jack and win twice the bet if the dealer does have black jack. Suppose you have $20 wagered and that you have a king and a 9 and the dealer has an ace. Should you buy insurance for $10?

###### Solution

We construct a probability distribution table

\(x\) | \(P(x)\) |

-10 | 34/49 |

20 | 15/49 |

(There are 49 cards that haven't been seen and 15 are 10JKQ (jacks, kings and queens) and the other 34 are non tens.)

We define the

expected value = \(S \times P(x)\)

We calculate:

\[-10(34/49) + 20(15/49) = -40/49 \nonumber \]

Hence the expected value is negative so that we should not buy insurance. What if I am playing with my wife. My cards are 2 and a 6 and my wife's are 7 and 4. Should I buy insurance? We have:

\(x\) | \(P(x)\ |

-10 | 31/47 |

20 | 16/47 |

We calculate:

\[-10(31/47) + 20(16/47) = 10/47 = 0.21 \nonumber \]

Hence my expected value is positive so that I should buy insurance.

## Standard Deviation

We compute the standard deviation for a probability distribution function the same way that we compute the standard deviation for a sample, except that after squaring \(x - m\), we multiply by \(P(x)\). Also we do not need to divide by \(n - 1\).

Consider the second insurance example:

\(x\) | \(P(x)\) | \(x - \overline{x}\) | \( (x - \overline{x}\)^2\) |

-10 | 31/47 | -10.21 | 104 |

20 | 16/47 | 19.79 | 392 |

Hence the variance is

\[104(31/47) + 392(16/47) = 202 \nonumber \]

and the standard deviation is the square root of the variance, which is14.2.

## Combining Distributions

If we have two distributions with independent random variables \(x\) and \(y\) and if \(a\) and \(b\) are constants then if

\( L = a + bx \) and \( W = ax + by\)

then

- \(m_L = a + bm\)
- \(\sigma_L^2 = b^2s^2\)
- \(\sigma_L = |b|\, s\)
- \(m_W = a\,m_x + b\,m_y \)
- \(\sigma_W^2 = a^2s_1^2 + b^2s_2^2 \)
- \( \sigma_W = \sqrt{a^2\, \sigma_x^2 + b^2\, \sigma_y^2}\)

Gamblers who played both black jack and craps were studied and it was found that the average amount of black playing per weekend was 7 hours with a standard deviation of 3 hours. The average amount of craps play was 4 hour with a standard deviation of 2 hours. What is the mean and standard deviation for the total amount of gaming?

###### Solution

Here \(a\) and \(b\) are 1 and 1. The mean is just

\[7 + 4 = 11 \nonumber \]

and the standard deviation is just

\[ \sqrt{3^2 + 2^2} = \sqrt{13} \nonumber \]

If each player spends about $100 per hour on black jack and $200 per hour on craps, what will be the mean and standard deviation for the amount of money that the casino wins per person?

###### Solution

Here a and b are 100 and 200. The mean is

\[100(7) + 200(4) = 1,500 \nonumber \]

and the standard deviation is

\[ \sqrt{(100^2)(3^2)+(200^2)(2^2)}=100\sqrt{17} \nonumber \]

If the players spend $150 on the hotel, find the mean and standard deviation of the total amount of money that the players spend.

Here

\[ L = 150 + x \nonumber \]

where \(x\) is the result from part B. Hence the mean is

\[ 150 + 1500 = 1,650 \nonumber \]

and the standard deviation is the same as part B since the coefficient is 1.

## The Binomial Distribution

There is a type of distribution that occurs so frequently that it has a special name. We call a distribution a binomial distribution if all of the following are true

- There are a fixed number of trials, \(n\), which are all independent.
- The outcomes are
*Boolean*, such as True or False, yes or no, success or failure. - The probability of success is the same for each trial.

For a binomial distribution with \(n\) trials with the probability of success \(p\) and failure \(q\), we have

\[ P(r \text { successes}) = C_{n,r}\, p^r \,q^{n-r} \nonumber \]

Suppose that each time you take a free throw shot, you have a 25% chance of making it. If you take 15 shots, what is the probability of making exactly 5 of them.

###### Solution

We have \( n = 15 \), \( r = 5\), \(p = 0.25 \), and \(q = 0.75\)

Compute

\[ C_{15,5}\, 0.25^5 \,0.75^{10} = 0.165 \nonumber \]

There is a 16.5 % chance of making exactly 5 shots.

What is the probability of making fewer than 3 shots?

###### Solution

The possible outcomes that will make this happen are 2 shots, 1 shot, and 0 shots. Since these are mutually exclusive, we can add these probabilities.

\[ C_{15,2} \, 0.25^2\, 0.75^{13} + C_{15,1}\, 0.25^1\, 0.75^{14} + C_{15,0}\, 0.25^0 \,0.75^{15} \nonumber \]

\[ = 0.156 + 0.067 + 0.013 = 0.236 \nonumber \]

There is a 24 % chance of sinking fewer than 3 shots.

### Contributors

- Larry Green (Lake Tahoe Community College)