Conditional Probability

• Conditional probability is the probability of some event A, assuming event B

• denoted by $P(A|B)$, which is read as "probability of A, given B"

The probability of getting a Jack given a face card (J, Q or K), is an example of event with conditional probability - its probability depends on the probability of picking a face card out of the deck in the first place.

Conditional probability is calculated using Bayes theorem:

$\displaystyle P(A|B) = \frac{P(B|A)\ P(A)}{P(B)}$

The probability of A, given B, is the ratio of the probabilities of A and B, multiplied by the probability of B, given A.

$\displaystyle P(A|B) = \frac{P(A)}{P(B)}\ P(B|A)$

Example 1

The probability of picking a face card (J,Q,K) out of the deck:

• 12 faces in 52 cards: P(f) = 12/52 = 0.23

Example 2

The probability of getting a Jack, given a face card:

• we need P(J|F), which depends on

• the probability of picking a face card: P(F) = 12/52

• the probability of picking a Jack: P(J) = 4/52

• the probability of getting a face card, given Jack, P(F|J) = 1

• The probability of getting a Jack, given a face card: 1/3

$$\displaystyle P(J|F) = \frac{P(J)}{P(F)} P(F|J) = \frac{4}{52} \div \frac{12}{52} = \frac{4}{52} \times \frac{52}{12} = \frac{4}{12} = \frac{1}{3}

Example 4

Prob of drawing:

• blue, blue: 1/10

• blue, red : 3/10

• red, blue : 3/10

• red, red : 3/10

• total: 1 = 10/10

Example 4

Prob that the second drawn marble is red (we ignore the first marble drawn)? We need to examine paths where the second drawn marble is red, (_, red). This implies aggregating the prob of these paths, i.e. using OR, so we add the probabilities:

• (blue, red) = 3/10

• (red, red) = 3/10

• so, the prob is 3/10 + 3/10 = 6/10 = 3/5