maths.free › Probability Theory › 10. Probability › Conditional Probability and Independent Events

Conditional Probability and Independent Events

A jar contains twenty marbles of which six are red, nine are blue and the remaining five are green.

Conditional Probability and Independent Events

A jar contains twenty marbles of which six are red, nine are blue and the remaining five are green. While blindfolded, Xing selects two of the twenty marbles random (without replacement) and puts one in his left pocket and one in his right pocket. He then takes off the blindfold.

The probability that the marble in his left pocket is red is \(6/20\). But Xing first reaches into his right pocket, takes this marble out and discovers that it is blue. Is the probability that the marble in his left pocket is red still \(6/20\)? Intuition says that it's slightly higher than that. Here's a more formal framework for answering such questions.

Let \((S,P)\) be a probability space and let \(B\) be an event for which \(P(B)>0\). Then for every event \(A\subseteq S\), we define the probability of \(A\), given \(B\), denoted \(P(A|B)\)\(P(A|B)\)probability of \(A\) given \(B\), by setting \(P(A|B)=P(A\cap B)/P(B)\).

Example

Consider the jar of twenty marbles from the preceding example. A second jar of marbles is introduced. This jar has eighteen marbles: nine red, five blue and four green. A jar is selected at random and from this jar, two marbles are chosen at random. What is the probability that both are green? Bob is on a roll. He says, Let \(G\) be the event that both marbles are green, and let \(J_1\) and \(J_2\) be the event that the marbles come from the first jar and the second jar, respectively. Then \(G= (G\cap J_1)\cup (G\cap J_2)\), and \((G\cap J_1)+(G\cap J_2)=\emptyset\). Furthermore, \(P(G|J_1)=\binom{5}{2}/\binom{20}{2}\) and \(P(G|J_2)=\binom{4}{2}/\binom{18}{2}\), while \(P(J_1)=P(J_2)=1/2\). Also \(P(G\cap J_i)=P(J_i)P(G|J_i)\) for each \(i=1,2\). Therefore, \[\begin{aligned}\end{aligned}\] That's about \(4.6\)%.

Now Alice is speechless.

Independent Events

Let \(A\) and \(B\) be events in a probability space \((S,P)\). We say \(A\) and \(B\) are independent if \(P(A\cap B)=P(A)P(B)\). Note that when \(P(B)\neq 0\), \(A\) and \(B\) are independent if and only if \(P(A)=P(A|B)\). Two events that are not independent are said to be dependent. Returning to our earlier example, the two events (\(A\): the marble in Xing's left pocket is red and \(B\): the marble in his right pocket is blue) are dependent.

Example

Consider the two jars of marbles from . One of the two jars is chosen at random and a single marble is drawn from that jar. Let \(A\) be the event that the second jar is chosen, and let \(B\) be the event that the marble chosen turns out to be green. Then \(P(A)=1/2\) and \(P(B)=\frac{1}{2}\frac{5}{20}+ \frac{1}{2}\frac{4}{18}\). On the other hand, \(P(A\cap B)=\frac{1}{2} \frac{4}{18}\), so \(P(A\cap B)\neq P(A)P(B)\), and the two events are not independent. Intuitively, this should be clear, since once you know that the marble is green, it is more likely that you actually chose the first jar.

Example

A pair of dice are rolled, one red and one blue. Let \(A\) be the event that the red die shows either a \(3\) or a \(5\), and let \(B\) be the event that you get doubles, , the red die and the blue die show the same number. Then \(P(A)=2/6\), \(P(B)=6/36\), and \(P(A\cap B) = 2/36\). So \(A\) and \(B\) are independent.

Symbols used here

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

imaginary unit
i² = −1.

\neq

not equal
The two sides are different.

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

E[X],\ \operatorname{Var}(X)

expected value, variance
Probability-weighted average of X; its spread.

N(\mu, \sigma^2),\ z

normal distribution, z-score
The bell curve with mean μ and variance σ²; (x − μ)/σ.

\mu(A),\ \sigma\text{-algebra}

measure of A
Size of a set; the family of sets that can be measured.

Questions people ask

What is the difference between probability and statistics?

Probability goes from a known model to what the data should look like; statistics goes from data back to the model. Probability theory is the deductive half.

What does the law of large numbers promise?

That the average of many independent samples converges to the expected value. It says nothing about any single trial.

Jiri gị onwe gị

Parts of this page are adapted from Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Oge Probability Theory

Sample spaces and the axioms Random variables and expectation The common distributions The law of large numbers and the central limit theorem