Probability of A and B

Definition

P(A and B) simply means "The probability of occurrence of both events, A, and B."

For example, if a coin is tossed and a number cube is rolled together, the probability of getting heads on the coin and the number 4 on the cube is represented by P(heads and 4).

P(A and B) is also written as `P(A \cap B)` and P(AB).

Formula

There are two formulas to find P(A and B) depending on whether A and B are independent or dependent events.

If events A and B are independent,

`P("A and B") = P(A) * P(B)`

If the probability of event A does not change whether event B occurs or not, and vice-versa, then A and B are called independent events. In such case, the probability of occurrence of both A and B is the product of their individual probabilities.

On the other hand, if events A and B are not independent,

`P("A and B") = P(A) * P(B|A)`

Events A and B are said to be dependent if the probability of either event changes depending on whether the other event has occurred or not. In such a case, the probability of occurrence of both A and B together is given by the above formula.

`P(B|A)` is the conditional probability of B given that event A has occurred. To understand it, read this post on conditional probability

In addition to the above formulas, there is also a method to find P(A and B) by simple counting techniques. This is shown in the example below.

Solved Examples

Problem 1

Two coins are tossed together. What is the probability that you get heads on the first coin and tails on the second.

First determine whether the two events are independent or dependent. Since the outcome of of tossing one coin does not affect the outcome of tossing the other coin, hence the two events are independent.

Find the probabilities of each event.

P(heads on first coin) = `1/2`
P(tails on second coin) = `1/2`

Now use the appropriate formula to calculate the probability of occurrence of both events. Since the two events are independent, we will use the following formula.

P(A and B) = P(A) * P(B)

P(heads on first and tails on second) = P(heads on first) * P(tails on second)

P(heads on first and tails on second) = `1/2 * 1/2 = 1/4`

Problem 2

A coin and a number cube is tossed together. What is the probability that you get heads on the coin and the number 3 on the number cube?

Determine whether the given events are dependent or independent. Since the outcome of tossing the coin does not affect the outcome of rolling the number cube in any form, therefore the two events are independent.

Find the individual probabilities of both events.
P(heads on coin) = `1/2`
P(3 on cube) = `"Number of 3's on a number cube"/"Total number of faces on a number cube" = 1/6`

Apply the appropriate formula to calculate the probability of both events occurring.

P(heads on coin and 3 on cube) = P(heads on coin) * P(3 on cube) = `1/2 * 1/6 = 1/12`

Therefore the probability of getting a heads on tossing a coin and 3 on rolling a number cube is `1/12`.

Problem 3

There are two bags of marbles. One bag contains three blue and four white marbles while the other bag contains two red and two green marbles. One marble is selected from each bag. What is the probability that you get a blue marble from the first bag and a red marble from the second?

Determine whether the two events are independent or dependent. Since the outcome of selecting a marble from the first bag does not affect the outcome of selecting a marble from the second bag, hence the two events are independent.

Find the individual probabilities of both events.
P(blue from first bag) = `"Number of blue marbles in the first bag"/"Total number of marbles in the first bag" = 3/7`
P(red from second) = `"Number of red marbles in the second bag"/"Total number of marbles in the second bag" = 2/4 = 1/2`
P(blue from first and red from second) = `3/7 * 1/2 = 3/14`

Problem 4

Two number cubes are tossed. What is the probability that you get an even number on the first cube and an odd number on the second?

The two given events are independent since the outcome of rolling the first cube does not affect that of rolling the second cube. Thus,

P(even on first and odd on second) = P(even on first) * P(odd on second)

There are a total of six numbers on a number cube out of which three are even and three are odd. Thus,

P(even on first) = `"Number of even numbers on a cube"/"Total number of numbers on a number cube" = 3/6 = 1/2`

P(odd on first) = `"Number of odd numbers on a cube"/"Total number of numbers on a number cube" = 3/6 = 1/2`

Applying the formula above,

P(even on first and odd on second) = `1/2 * 1/2 = 1/4`

Problem 5 

From a standard deck of 52 cards, you pick two cards. What is the probability that the first card is a red Ace and the second is a black Queen?

Before drawing the first card, there are a total of 52 cards and a total of two red aces in it. Thus,

P(red ace) = `2/52 = 1/26`

If the first card is a red ace, then there are a total of 51 and two black queens left in the deck. Thus,

P(black queen) = `1/51`

Since the two events above are independent, therefore,

P(red ace and black queen) = P(red ace) * P(black queen) = `1/25 * 1/51 = 1/1326`

Therefore the probability of getting a red ace and then a black queen on drawing two cards from a deck of 52 cards is `1/1326`