In this Lesson, we will learn some of the very popular, widely used formulas on Probability and solve numerous questions based on them.

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:__

1. 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

2. 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}

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}

__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)

= 1 – [P (Â) × P (Bˆ)]

= 1–[(1/4) × (1/6)]

= 1 – 1/24 = 23/24

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:__

1. On tossing two coins, Heads on one coin and Tails on the other coin.

2. On tossing two unbiased dice, an odd number on one dice and an even number on the other.

3. 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 1^{st} and T on 2^{nd}) = P (H on 1^{st}) × P (T on 2^{nd})

= ^{1}/_{2} × ^{1}/_{2} = ^{1}/_{4}

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

1. Both pass in the test.

2. At least one will pass in the test.

3. Neither A nor B will pass in the test.

4. 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.

1. 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}]

2. P (at least one will pass) = 1 – P (None of A, B will pass)

= 1 – P (Neither A will pass nor B will pass)

= 1 – P (A will not pass) × P (B will not pass)

= 1 – [(^{1}/_{3}) × (^{1}/_{5})] = 1 – ^{1}/_{15} = ^{14}/_{15}

3. P (Neither A nor B will pass) =

P (A will not pass) × P (B will not pass) =

(^{1}/_{3}) × (^{1}/_{5}) = ^{1}/_{15}

4. P (One will pass in the test) =

P (A will pass but not B) or P (A will not pass but B will)

= [P (A will pass) × P (B will not pass)] + [P (A will not pass) × P (B will pass)]

= [(^{2}/_{3}) × (^{1}/_{5})] + [(^{1}/_{3}) × (^{4}/_{5})]

= ^{2}/_{15} + ^{4}/_{15} = ^{6}/_{15} = ^{2}/_{5}

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

Examples:

1. The events of Heads and Tails on tossing a fair coin.

2. 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

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