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# Probability Distribution Tables

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## Probability Distribution Tables

** What is a discrete random variable? **

A random variable is a variable which takes numerical values and whose value depends on the outcome of an experiment. It is discrete if it can only take certain values.

**Example:**

Suppose you roll a die and let X be the number on the uppermost face. Then the values that can be obtained are discrete random variables.

x = 1, 2, 3, 4, 5 or 6

The probability of obtaining a 1 can be written as P(X = 1) = 1/6

The probability of obtaining a 2 can be written as P(X = 2) = 1/6

And so on for all the values of x, P(X = x).

If we list all these probabilities into a table, we get the **probability distribution of X**.

x | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|

P(X = x) | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 |

**Note:** Capital letters are used to denote the random variables, whereas lower case letters are used to denote the values that can be obtained.

In the above example we covered every possible event when rolling a die. We must get one of the six events listed in the table. These events are **exclusive events** and **always sum to 1.**

∑ P(X = x) = 1

**Example:**

Suppose you flip 2 coins and let X be the discrete random variable of the number of heads obtained. Write down the **probability distribution** of X.

First we need to know what values of x can be obtained.

Clearly, x = 0, 1 or 2, as we can either get no heads, 1 head or 2 heads.

Our next step is to calculate the probability of each (if you have trouble with this then go and have a look at the * Probability* topic).

P(X = 0) = P(tail and tail) = ½ × ½ = ¼

P(X = 1) = P(tail and head) or P(head and tail) = ½ × ½ + ½ × ½ = ½

P(X = 2) = P(head and head) = ½ × ½ = ¼

We now put these results into a table for the probability distribution of X...

x | 0 | 1 | 2 |

P(X = x) | ¼ | ½ | ¼ |

**Note:** The better you are at calculating probabilities, the quicker and easier these problems become. You should be able to write down the probability distribution of a discrete random variable with minimal workings.

Sometimes we are given a formula to calculate probabilities. We call this the **probability density function of X** or the **p.d.f. of X**.

**Example:**

The probability density function of a discrete random variable X is given by...

P(X = x) = kx^{3}

For x = 0, 1 or 2 where k is a constant.

Find the probability distribution of X.

P(X = 0) = k × 0^{3} = 0

P(X = 1) = k × 1^{3} = k

P(X = 2) = k × 2^{3} = 8k

So:

x | 0 | 1 | 2 |

P(X =x) | 0 | k | 8k |

From earlier, we know:

∑ P(X = x) = 1

(**Remember:** ∑ means 'sum of')

Therefore, 0 + k + 8k = 1

9k = 1

Hence k = 1/9

So the probability distribution of X is:

x | 0 | 1 | 2 |

P(X = x) | 0 | 1/9 | 8/9 |

**'Cumulative'** gives us a kind of running total so a cumulative distribution function gives us a running total of probabilities within our probability table.

The cumulative distribution function, F(x) of X is defined as:

F(x) = P(X ≤ x)

**Example:**

The probability distribution of a discrete random variable, X, is given by:

x | 1 | 2 | 3 | 4 | 5 |

F(x) | 0.2 | 0.43 | 0.62 | 0.8 | 1 |

Reading the table we see that:

P(X ≤ 3) = 0.62

P(X ≤ 2) = 0.43

This means to calculate a single probability we proceed as follows:

P(X = 3) | = P(X ≤ 3) − P(X ≤ 2) |

= 0.62 − 0.43 | |

= 0.19 |

For probabilities of **'greater than'**:

P(X >2) | = 1 - F(2) |

= 1 − P(X ≤ 2) | |

= 1 − 0.43 | |

= 0.57 |

* Question:* Try for yourself

For the above example, see if you can work out the following: