Joint probability and conditional probablity

Joint probability

• Joint probability is the probability of the occurence of 2 events simultaneously.
• 2 events are independent of each other in Joint probability.
• $$P(A \cap B) = P(A) \cdot P(B)$$

Examples

Ques. What is the probability of you getting an 8 and a red card while drawing a card from a deck?
Soln.

$$\begin{align*} & P(8 \cap red) = P(8) \cdot P(red) \\ & P(8 \cap red) = \frac{4}{52} \cdot \frac{26}{52} = \frac{1}{26} \end{align*}$$

Ques. A bag contains 3 red balls, 5 green balls, and 10 black balls. What is the probability of you getting either a red ball or a green ball when you randomly draw a ball from the bag?
Soln.

$$\begin{align*} & P(red) = \frac{3}{18} \\ & P(green) = \frac{5}{18} \\ & P(\text{red or green}) = \frac{3}{18} + \frac{5}{18} = \frac{4}{9} \end{align*}$$

Conditional probability

• It is the probability of an event when some other event has already occurred.
• In Conditional probability, 2 events are dependent on each other.
• Notation: $P(B \vert A)$, read as Probability of B given that A has already occurred

Formula

$$P(B \vert A) = \frac{P(B \cap A)}{P(A)}$$

Examples

Ques. In a game of cards, a player wins if he draws two cards from the same unit. If the first card is a spade, what is the probability that the player will win?
Soln. The probability of another spade from the deck would be $\dfrac{12}{51}$. Since 1 spade has already been drawn from the deck as first event.

Ques. A bag contains 4 red balls and 5 green balls. You draw a ball from it without replacing it with another one. What is the probability of you drawing a green ball in the second draw, provided that it was a red ball that you got in the first draw?
Soln. $\text{The probability of red ball} = \dfrac{4}{9}$. Now the probability of ball being green in second draw $= \dfrac{5}{8}$.

Baye’s Theorem

This describes the probability of an event, based on prior knowledge of conditions that might be related to the event. If you know the Conditional probability $P(A \vert B)$, you can use Baye’s rule to find out the reverse probability $P(B \vert A)$.

$$P(A \vert B) = \frac{P(B \vert A) \cdot P(A)}{P(B)}$$

Generic formula

$$P(A_i \vert B) = \frac{P(B \vert A_i) \cdot P(A_i)}{\sum_{i=1}^{n}{P(B \vert A_i) \cdot P(A_i)}}$$