# What is the probability of either A or B or both of the two events happening?

Another word for probability is a possibility. Probability is a branch of mathematics, concerning how likely an event is to occur. The probability of number is indicated from zero to one. In mathematics, Probability has been described to predict how likely there are chances of occurring a likely event. The meaning of probability is the chance to which something is probably or certainly happen. Given below are the terminologies used in probability,

Set Language | Set Notation |

Subset A (or event A) | eg, A |

Complement of A | A^{c} |

Union of A and B | A ∪ B |

Intersection of A and B | A ∩ B or AB |

A and B are disjoint (mutually exclusive) | P(A ∩ B) = 0. |

A is a subset of B | A ⊆ B |

### Probability Rules

There are different probability rules like a complementary rule, difference rule, inclusion-exclusion rule, conditional probability, etc. Let’s take a look at these rules in detail,

**Complement Rule:**According to this rule, the possibility that A does not occur is equal to the possibility that the complement of event A occurs.

Formula ⇒ P(A^{c}) = 1 – P(A).

**Difference Rule:**According to this formula if A is a subset of B, then the possibility of B occurring but not A is,

⇒ P(B) – P(A) = P(B A^c).

**Inclusion-Exclusion Rule:**According to this rule, the possibility of either A or B (or both) occurring is,

⇒ P(A U B) = P(A) + P(B) – P(AB).

**Conditional Probability:**According to this probability the measure of the probability of an event occurring given that another event has already occurred = P(A|B). In other words, among those instance where B has occurred, P(A|B) is the proportion of cases in which event A occurs.**Multiplication Rule:**According to this rule, the possibility of both A and B occurring is equal to the possibility that B occurs times the conditional possibility that A occurs given that B occurs.

⇒ P(AB) = P(B) P(A|B).

Consequently, the conditional probability is given by P(A|B) = P(AB)/P(B). Similarly, the possibility that A occurs times the conditional possibility that B occurs given that A has: P(A) P(B|A) = P(AB), so, P(B|A) = P(AB)/P(A).

**Bayes’ Rule:**This formula relates the depending probability of B given A to the depending probability of A given B.

⇒ P(B|A) = P(B) P(A|B) / P(A)

**Average Formula:**Say that the set A can be completely separated into n combined exclusive subsets. Then the all-inclusive probability of A is equal to the average probability of A in the subgroup weighted by the probability of those subgroup:

⇒ P(A) = P(A|B_{1}) P(B_{1}) + P(A|B_{2}) P(B_{2}) + … + P(A|B_{n}) P(B_{n})

**Independence:**According to this rule if the possibility of A does not depend on whether or not B occurs, then we say that A and B are independent which means they are not dependent on each other.

For independent events ONLY,

⇒ P(A|B) = P(A)

⇒ P(AB) = P(A) P(B)

### What is the probability of either A or B or both of the two events happening?

**Solution:**

Let’s consider A and B are the likely happening event. According to Inclusion-Exclusion Rule:

The probability of either A or B (or both) occurring is,

⇒ P(A U B) = P(A) + P(B) – P(AB).

For example: If a coin is tossed two times what is the probability of getting either head or tail or both tails.

When a coin is tossed, either a HEAD or a TAIL is obtained.

The Probability of either is the same, which is 0.5 or 1⁄2.

When two coins flipped the possible outcome are:

Flip 1Flip 2H H H T T H T T Note that there is an equal possibility of happening any of the four combinations as

P(HEADS) = P(TAILS) = 0.5

There are 4 possible outcomes i.e., (H, H), (H, T), (T, H), and (T, T). Only one option is of the four is (TAILS, TAILS), so P(FLIP 1 and FLIP 2 = TAILS, TAILS).

= (1/4) = 0.25 = 25%

Now apply the formula: The probability of either A or B (or both)events occurring is

⇒ P(A U B) = P(A) + P(B) – P(AB).

= 1⁄2 + 1⁄2 – 1⁄4

= 2⁄2 – 1⁄4

= 0.75

### Similar Problems

**Question1: If the probability of having green eyes is 10%, the probability of having brown hair is 75%, and the probability of being a green-eyed brown-haired person is 9%, let us assume, A as green eyes and B as brown hair, what is the probability of:**

**Not having green eyes?****Have green eyes but not brown hair?****Have green eyes and/or brown hair?**

**Solution:**

Not have green eyes?Formula: P(A

^{c})= 1 – P(A) (According to Complement Rule)

= 1 – 10%

= 0.9 or 90%

Having green eyes but not brown hair?Formula: P(A) – P(AB)

= 10% – 9%

=0.091 or 9.1%

Having green eyes and/or brown hair?Formula: P(A U B)

= P(A) + P(B) – P(AB) (According to inclusion-exclusion rule)

= 10% + 75% – 9%

= 0.15925 or 15.925%

**Question 2:If the probability of having black shoes is 9%, the probability of having a brown shirt is 75%, and the probability of a person wearing a black-shoes brown shirt is 10%, let us consider, A as black shoes and B as a brown shirt, what is the probability of:**

**Not having black shoes?****Have black shoes but not a brown shirt?****Have black shoes and/or a brown shirt?**

**Solution:**

Not having black shoes?Formula: P(A

^{c})= 1 – P(A) (According to Complement Rule)

= 1- 9%

= 0.91 or 91%

Having black shoes but not brown shirt?Formula: P(A) – P(AB)

= 9% – 10%

= 0.081 or 8.1%

Having black shoes and/or brown shirt?Formula: P(A U B)

= P(A) + P(B) – P(AB) (According to inclusion-exclusion rule)

= 9% + 75% – 10%

= 0.14175 or 14.175%