introduction o robabilityProbability Addition Rule

The Addition Rule of Probability Definition & Examples worksheet.Probability is one of the most important topics of the mathematics. Probability of an certain event, say AA is symbolized by P(A)P(A). Always it lies between 0 and 1.

We define probability as the number of outcomes of an event associated with a random experiment divided by the total number of outcomes possible in that experiment. Let n(S)n(S) denote the number of outcomes of the experiment and A be an event whose outcomes are n(A)n(A).
Then, probability of occurrence of event A is given by: P(A)P(A) = n(A)n(S)n(A)n(S). In this topic we will learn about probability addition formula in detail with couple of solved examples.

Formula

Suppose we have two events X and Y associated with a random experiment E. then the probability of occurrence of either X or Y or both is given by:

P(X or Y)P(X or Y) = P(X)+P(Y)P(X)+P(Y) – P(X andY)P(X andY)

→→ P(X∪Y)P(X∪Y) = P(X)+P(Y)P(X)+P(Y) – P(X∩Y)P(X∩Y)

Where P(X)P(X) is the probability of occurrence of event X,P(Y)X,P(Y) is the probability of occurrence of event YY and P(X and Y)P(X and Y) [also written as P(X ∩Y)P(X ∩Y) is the probability of occurrence of both XX and YY events simultaneously.

We have two cases:

Case I:

When the two events XX and YY are not mutually exclusive that is the occurrence of event XX affects the probability occurrence of event YY or vice versa. In this case the intersection probability that is P(X ∩ Y)P(X ∩ Y) ≠≠ 0. And we have,
P(X or Y)P(X or Y) = P(X)+P(Y)–P(X∩Y)P(X)+P(Y)–P(X∩Y)

Case II:

When the two events XX and YY are mutually exclusive that is the occurrence of event XX does not affects the probability occurrence of event Y or vice versa. In this case the intersection probability that is P(X∩Y)P(X∩Y) = 0. And we have,

P(X or Y)P(X or Y) = P(X)P(X) + P(Y)

Problems

Some examples on probability addition rule are illustrated below:

Example 1:
We have rolled a six sided die. Find the probability of getting an even number or a 5.

Solution:

Let E be the experiment of rolling a die and S be the sample space, then

S = {2, 4, 1, 5, 3, 6} => n (S) = 6

A = event of getting an even number = {4, 2, 6} => n (A) = 3

B = event of getting a 5 = {5} => n (B) = 1

So we have P (A) = 3636, P (B) = 1616.

It is clear that the two events are mutually exclusive, hence P(A∩B)P(A∩B) = 0

So according to the probability addition rule we have:

$P (A\ or\ B) = 3636 + 1616 = 4646 = 2323

Example 2: We are given a deck of 52 cards. We pick a card from it. Find the probability of getting a heart or a red king.

Solution: Let S be the sample space then n(S)n(S) = 52.

AA = event of getting a red king => n(A)n(A) = 2 => P(A)P(A) = 252252

BB = event of getting a heart => n(B)n(B) = 13 => P(B)P(B) = 13521352

Clearly AA and BB are not mutually exclusive as a red king can also be of heart.

=> n(A ∩ B)n(A ∩ B) = 1 => P(A∩B)P(A∩B) = 152152.

By addition rule of probability we have:

P (A or B) = 252252 + 13521352 – 152152 = 14521452 = 726726

Example 3: What is the probability of getting a total of 3 or 7, when two dice are rolled?
Solution: Let event A be the total of 7 and event B the total of 3.

When we roll two dice we have 36 events in a sample space.

Number of events getting total of 3 = 2

Number of events getting total of 7 = 6

=> P(A)P(A) = 636636 and P(B)P(B) = 236236

Both the events are mutually exclusive since they do not have any common event.

i.e. P(A∩B)P(A∩B) = 0

Now P(A∪B)P(A∪B) = P(A)P(A) + P(B)P(B) = 636636 + 236236 = 836836 = 2929.

 

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