4.1.1 Discrete Probability Distributions

Discrete Random Variables

A discrete random variable is a quantity that can randomly take a discrete (not continuous) set of values.

e.g: $X$ is the outcomes possible on a fair die.
$X$ is a discrete random variable because its value is assigned by chance and there are a number of discrete outcomes $\{1, 2, 3, 4, 5, 6\}$ . The distribution for $X$ is a list of probabilities and associated outcomes:

\boxed {\begin{array}{c|c|c|c|c|c|c} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(X = x) & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \end{array} }

(Referred to as "distribution of $X$ ")

Small $x$ for particular outcome.
Capital $X$ for the value of the variable.

A probability distribution for a function could be given as a formula:

P(X = x) = \begin{cases} 2kx & 1 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}

It is highly desirable to put such distribution functions in table form:

\boxed {\begin{array}{c|c|c|c|c} x & 1 & 2 & 3 & 4 \\ \hline P(X = x) & 2k & 4k & 6k & 8k \end{array}}

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All probabilities sum to 1. It is possible to calculate $k$ using $2k + 4k + 6k + 8k = 1$ .

lightbulbExample

Example Q1 (OCR 4766, Jun 2016, Q4) [Modified] The probability distribution of the random variable X is given by the formula:

P(X = r) = \frac{k}{r(r-1)} \quad \text{for} \, r = 2, 3, 4, 5, 6.

Question: Show that the value of k is 1.2. Using this value of k, show the probability distribution of $X$ in a table.

Step 1: Set up the Total Probability Equation

Since the total probability must sum to 1, we write

P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) = 1

Substitute the formula for each P(X = r)

\frac{k}{2} + \frac{k}{6} + \frac{k}{12} + \frac{k}{20} + \frac{k}{30} = 1

You can also show this in a table:

\boxed {\begin{array}{c|c|c|c|c|c} r & 2 & 3 & 4 & 5 & 6 \\ \hline P(X = r) & \frac{k}{2} & \frac{k}{6} & \frac{k}{12} & \frac{k}{20} & \frac{k}{30} \end{array} }

Step 2: Simplify and Solve for k

Combine the fractions by finding a common denominator

\Rightarrow \frac{5k}{6} = 1

Solve for k:

\quad \Rightarrow k = \frac{6}{5} = :success[1.2]

Step 3: Calculate the Probabilities

Using k = 1.2, calculate the probabilities for each value of r:

P(X=2)=\frac{1.2}{2}=:success[0.6]

P(X=3)=\frac{1.2}{6}=:success[0.2]

P(X=4)=\frac{1.2}{12}=:success[0.1]

P(X=5)=\frac{1.2}{20}=:success[0.06]

P(X=6)=\frac{1.2}{30}=:success[0.04]

Put this into a table:

\boxed {\begin{array}{c|c|c|c|c|c} r & 2 & 3 & 4 & 5 & 6 \\ \hline P(X = r) & 0.6 & 0.2 & 0.1 & 0.06 & 0.04 \end{array} }

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Example: Calculate $P(X>3)$ using the table below Using the above table, perform the following calculations:

\boxed {\begin{array}{c|c|c|c|c|c|c} r & 2 & 3 & 4 & 5 & 6 \\ \hline P(X = r) & 0.6 & 0.2 & 0.1 & 0.06 & 0.04 \end{array}}

Step 1: Identify the relevant probabilities

We need to calculate the probability that the random variable X is greater than 3.

Therefore we need the probabilities for $X=4$ , $X=5$ and $X=6$

From the table we can see these probabilities:

P(X=4)=0.1

P(X=5)=0.06

P(X=6)=0.04

Step 2: Add the probabilities

P(X > 3) = 0.1 + 0.06 + 0.04

P(X>3)= :success[0.2]

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Example: Calculate $P(4 \leq X < 6)$ using the table below

\boxed {\begin{array}{c|c|c|c|c|c|c} r & 2 & 3 & 4 & 5 & 6 \\ \hline P(X = r) & 0.6 & 0.2 & 0.1 & 0.06 & 0.04 \end{array}}

Step 1: Identify the relevant probabilities

We need to calculate the probability that X is between 4 and 6, inclusive of 4 ( $\leq$ ) but excluding 6 ( $<$ ).

This means we are interested in the probabilities for $X=4$ and $X=5$

From the table we can see these probabilities:

P(X=4)=0.1

P(X=5)=0.06

Step 2: Add the probabilities

P(4 \leq X < 6) = 0.1 + 0.06

P(4 \leq X < 6) = :success[0.16]

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Example: Calculate $P(2.5 \leq X \leq 4.6)$ using the table below

\boxed {\begin{array}{c|c|c|c|c|c|c} r & 2 & 3 & 4 & 5 & 6 \\ \hline P(X = r) & 0.6 & 0.2 & 0.1 & 0.06 & 0.04 \end{array}}

Step 1: Identify the relevant probabilities

We need to calculate the probability that $X$ is between 2.5 and 4.6, inclusive of both 3 and 4 since $X$ is a discrete random variable.

This means we are interested in the probabilities for $X=3$ and $X=4$

From the table we can see these probabilities:

P(X=3)=0.2

P(X=4)=0.1

Step 2: Add the probabilities

P(2.5 \leq X \leq 4.6) = 0.2 + 0.1

P(2.5 \leq X \leq 4.6)= :success[0.3]

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Example: Combined Events Say $X$ has the distribution as was previously:

\boxed {\begin{array}{c|c|c|c|c|c|c} r & 2 & 3 & 4 & 5 & 6 \\ \hline P(X = r) & 0.6 & 0.2 & 0.1 & 0.06 & 0.04 \end{array}}

Consider the situation where two independent values of $X$ are sampled: $X_1$ and $X_2$ .

Question: Find $P(X_1 = X_2)$

Step 1: Understand the Question

We are asked to find the probability that two independent values, $X_1$ and $X_2$ , sampled from the same probability distribution are equal.

This means we are looking for the probability that $X_1 = X_2$ , where both values come from the given distribution of $X$ .

Step 2: Identify the Relevant Outcomes

The values of $X_1$ and $X_2$ can both be one of the following: 2, 3, 4, 5 or 6.

We are interested in the probability that $X_1$ equals $X_2$ for each of these outcomes.

These are the cases where $X_1 = X_2$ for each value of $r$ .

This can be demonstrated in this table:

\boxed {\begin{array}{c|c|c|c|c|c|c} X_1, & 2 & 3 & 4 & 5 & 6 \\ X_2, & 2 & 3 & 4 & 5 & 6 \end{array}}

Step 3: Multiply the Probabilities

Since the two values $X_1$ and $X_2$ are independent, the probability that $X_1=X_2$ for each value of r is the product of the individual probabilities $P(X=r)$ .

For each possible value of r, the corresponding probabilities are:

P(X_1=X_2=2)=0.6^2

P(X_1=X_2=3)=0.2^2

P(X_1=X_2=4)=0.1^2

P(X_1=X_2=5)=0.06^2

P(X_1=X_2=6)=0.04^2

Step 4: Add the Probabilities

P(X_1 = X_2) = 0.6^2 + 0.2^2 + 0.1^2 + 0.06^2 + 0.04^2

P(X_1 = X_2)= :success[0.4152]

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Example: Probability of each outcome Say $X$ has the distribution as was previously:

\boxed {\begin{array}{c|c|c|c|c|c|c} r & 2 & 3 & 4 & 5 & 6 \\ \hline P(X = r) & 0.6 & 0.2 & 0.1 & 0.06 & 0.04 \end{array}}

Consider the situation where two independent values of $X$ are sampled: $X_1$ and $X_2$ .

Question: Find $P(X_1 + X_2 = 5)$

Step 1: Understand the Question

We are asked to find the probability that the sum of two independent values, $X_1$ and $X_2$ , equals 5.

We need to find all the combinations of $X_1$ and $X_2$ that satisfy this condition.

Step 2: Identify the Relevant Outcomes

From the table, the combinations of $X_1$ and $X_2$ that sum to 5 are:

X_1=2\ \text{and}\ X_2=3

X_1=3\ \text{and}\ X_2=2

Step 3: Multiply the Probabilities for Each Combination

Since $X_1$ and $X_2$ are independent, the probability of each outcome is the product of the individual probabilities:

P(X_1=2\ \text{and}\ X_2=3) = 0.6 \times 0.2 = 0.12

P(X_1=3\ \text{and}\ X_2=2) = 0.2 \times 0.6 = 0.12

Step 4: Add the Probabilities

P(X_1 + X_2 = 5) = 0.12+0.12

P(X_1 + X_2 = 5)=:success[0.24]

Discrete Probability Distributions Simplified Revision Notes for A-Level AQA Maths Statistics

4.1.1 Discrete Probability Distributions

Discrete Random Variables

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