**In this Lesson, we will learn some of the very popular, widely used formulas on Probability and solve numerous questions based on them: P (E) +P (Ê) = 1**

Let E denote an Event happens and Ê, the Event E will not happen. Let P (E) denote the probability the event E happens and P (Ê), the probability the Event E does not happen. Then:

*P (E) + P (Ê) = 1*

And also:

*P (Ê) = 1 – P(E)*

## Examples:

- On tossing a coin, either Heads or Tails will appear. Let E denote Heads. Then, Ê will denote Not Heads, same as Tails. In a single unbiased (fair) coin, P (H) + P (T) = 1. Replace H by E and T by Ê, so that we get: P (E) + P (Ê) = 1
- Toss a dice. What is the probability that a multiple of 3 does not appear? Solution: On a dice, the Multiples of 3 are: 3 and 6. Therefore, P (multiple of 3) = 2/6 = 1/3. Now, probability the number on the dice is not a multiple of 3 is: 1 – 1/3 = 2/3
- Probability a student will not pass a test is 2/5. What is the probability he will pass the test? Solution: P (will pass) = 1– P (will not pass) = 1 – 2/5 = 3/5
- P (at least one) = 1– P(None)

### Example 1:

Toss two coins. What is the probability that at least one heads will appear?

#### Solution:

P (E) = (Number of favorable outcomes)/ (Total outcomes)

Total outcomes on tossing two coins are 4, as below:

HH, HT, TH, TT

In 4 outcomes, 1 outcome, i.e. TT gives no Heads.

P (No heads) = 1/4

Therefore, P (at least one heads) = 1 – 1/4 = 3/4

Alternative (but not useful):

The event at least one heads denotes one or more heads.

So, probability of 1 heads or 2 heads is:

P (1 Heads) + P (2 Heads) = 2/4 + 1/4 = 3/4

(Alternative method is necessary when we find probability of at least two or more)

### Example 2:

Two students A and B attempt to solve a question. What is the probability the question is solved if the probability A solves is 2/3 and the probability B solves is 4/5?

#### Solution:

Each of A, B will solve or not solve the question without affecting the probability of the other (to solve or not solve)

Any one of A, B needs to solve for the question to be solved.

So, the required event E is at least one of A, B solves the question.

P (at least one solves) =1 –P(no one solves)

P (A) =2/3, P (Â) = 1– 2/3 = 1/3

P (B) =4/5, P (Bˆ) = 1 – 4/5 = 1/5

P (at least one solves) =

1 –P (no one solves) =1 – [(1/3) × (1/5)] = 1– (1/15) = 14/15

### Example 3:

Two shooters A and B fire shots at a target. What is the probability at least one hits the target, if P (A) = 3/4 and P (B) = 5/6?

#### Solution:

P (at least one will hit) = 1– P (No one will hit the target)

Independent Events:Two events A and B are said to be independent events, if the probability of occurrence of each event does not affect the probability of occurrence of the other.

Examples of Independent Events:

- On tossing two coins, Heads on one coin and Tails on the other coin.
- On tossing two unbiased dice, an odd number on one dice and an even number on the other.
- The result of two students taking a test.If two events A and B are independent, then the probability A and B will happen at the same time (joint occurrence of A and B or simultaneous occurrence of A and B) is: P (A and B) = P (A) × P (B). (The above formula is defined as multiplication rule of probability for joint occurrence of two independent events)

## Examples:

### Example 1.

Two coins are tossed. What is the probability Heads appears on first coin and Tails on the second?

#### Solution:

First use probability definition to solve:

When two coins are tossed, the 4 outcomes are: HH, HT, TH, and TT

1 outcome gives H on 1st coin and T on 2nd coin.

P (H on 1st and T on 2nd) = 1/4

Now, use Independent events formula to find:

H on 1st coin and T on 2nd are independent events as they do not affect each other.

P (H on 1st and T on 2nd) = P (H on 1st) × P (T on 2nd)

### Example 2:

Two students A and B take a test. The probability A passes in the test is 2/3 and B fails is 3/5. What is the probability that

- Both pass in the test.
- At least one will pass in the test.
- Neither A nor B will pass in the test.
- One will pass in the test.

#### Solution:

The event A passes or fails in the test will not affect or be affected by the event B passes or fails in the test. The two events are independent events.

- A and B stands for both A and B pass in the test. P (A and B) =
*P (A) × P (B) = (2/3) × (2/5) = 4/15. [P (B passes) = 1– P (B does not pass) = 1 – 3/5 = 2/5]* - P (at least one will pass) = 1 – P (None of A, B will pass)
- P (Neither A nor B will pass) = P (A will not pass) × P (B will not pass) =(1/3) × (1/5) = 1/15
- P (One will pass in the test) = P (A will pass but not B) or P (A will not pass but B will)

**Mutually Exclusive Events:**

Two events A and B which happen preventing (excluding) the other from happening at the same time are called mutually exclusive events.

**Examples:**

- The events of Heads and Tails on tossing a fair coin.
- The events of Numbers on tossing a dice.

**Note:** If two events A and B are mutually exclusive, then the probability they will occur at the same time is 0, i.e. P (A and B) = 0