## Probability: Operations

### Addition theorem for two events

If a and B are two events associated with the same random experiment (i.e. both the events from the same sample space S) then

i.e. p(A or B) = p(A) + p(B) - p(A and B)

**Solution:**

= 16 / 52

= 4 / 13

### Addition theorem for three events

Number of multiples of 3 = 33

Number of multiples of 4 = 25

Number of multiples of 5 = 20

Total = 78

Number of, Multiple of 3 and 4 = multiple of 12 = 8

Multiple of 4 and 5 = multiple of 20 = 5

Multiple of 3 and 5 = multiple of 15 = 6

Total = 19

Number of multiples of 3, 4 and 5 = multiples of 60 = 1

p = 78 / 100 - 19 / 100 + 1 / 100

= 60 / 100

= 3 / 5

### Probability of difference of two events

If A and B are two events associated with the same random experiment, then

p(A - B) = p(A) - p(A ∩ B)

p(B - A) = p(B) - p(A ∩ B)

p(A - B) = p(A) - p(A ∩ B)

p(A - B) = p(B) - p(A ∩ B)

### Probability of exactly one out of two events occurs

If A and B are two events associated with the same random experiment.

p(B-A) = p(B) - p(A ∩ B)

p[(A-B) or (B-A)] = p(A) + p(B) - 2p(A ∩ B)

Or p(A ∪ B) - p(A ∩ B)

### Probability of exactly one of the three events occurs

If A, B and C are three events associated with the same random experiment then

Additional Results

- p[exactly two of A,B and C occur] = p(A ∩ B) + p(B ∩ C) + p(A ∩ C) - 3p(A ∩ B ∩ C)
- p[at least two of A,B and C occur] = p(A ∩ B) + p(B ∩ C) + p(A ∩ C) - 2p(A ∩ B ∩ C)

## Advanced Probability Concepts

Now we will look into the advanced application of probability concepts. Sometimes it is quiet difficult to understand whether the question required an advanced kind of application such as 'Conditional Probability', so an aspirant should aware about the exact situations in which these advanced results are applicable.

The advanced probability concepts we are going to look into are

- Conditional probability
- Independent events
- Law of total probability
- Baye's theorem

### Conditional Probability

Let A and B are any two events associated with a random experiment. The probability of the occurrence of event A when the event b has already occurred is called the "conditional probability of the occurrence of A when B is given" and is denoted as P(A/B) read as "Probability of A given B".

The conditional probability P(A/B) is meaningful only when P(B) ≠ 0, i.e. when is not an impossible events.

Eg: Let us consider the experiment of flipping two coins simultaneously.

Let, A = {(H, H), (H, T)}

B = {(H, H), (T, H)}

C = {(H, H), (H, T), (T, H)}

P(A) = 2/4 = 1/2

P(B) = 1/2

P(C) = 3/4

A/B is the occurrence of the event A with condition that the event B already occurred.

i.e. P(A/B) = P(A ∩ B) / P(B)

Similarly, P(B/C) = 2/3

and P(A/C) = 2/3

P(A ∩ B) = P(A).P(A/B) or P(B). P(A/B)

### Probability - Independent Event

If the occurrence of one event doesn't depend upon the occurrence of another event, then both the events are called independent events.

Consider an urn, which contains 5 red balls and 4 green balls.

Let A1 - an event of selecting one red ball in first draw.

Let A2 - an event of selecting one green ball in second draw.

If the selection of balls one after another with replacement, then the occurrence of the event A1 can't after the occurrence of the event A2.

There for, A1 and A2 are two independent events.

If the selection of balls one after another without replacement, then the occurrence of A1 will after the occurrence of A2.

A1 and A2 are not two independent events.

There for, Two events A and B associated with a random experiment are said to be independent if,

There for, p(A/B) = p(A) and p(B/A) = p(B)

In general, if A1, A2, A3 , ....., An are independent events associated with a random experiment then

If A1, A2, A3 ..... An are 'n' independent events associated with the same random experiment, then

### Probability : Baye's Theorem

If E1,E2 ..... En are mutually exclusive and exhaustive with the same random experiment with non zero probabilities,

then for any event A which is a subset of total sample space [S= E1 ∪ E2 ... ∪ En] such that p(A) > 0, then

_{j = 1 to n}P(Ej).P(A/Ej)

**Example:**

There are two bags contains a certain number of gold coins and silver coins. 3 gold coins and 5 silver coins in Bag1, 6 gold coins and 2 silver coins in Bag2.A coin is drawn from one of the two bags, the coin drawn is gold. What is the probability that it has drawn from Bag2?

**Solution:**

Let A and B are the events of selecting Bag1 and Bag2 respectively.

p(A) = 1/2

p(B) = 1/2

Let C be the event of selecting a gold coin.

Then, p(C/A) = p(selecting a gold coin from Bag1) = 3/8

Similarly p(C/B) = 6/8

Coin selected is gold, then the probability of the selection done from Bag2 is p(B/C)

i.e. p(B/C) = p(B).p(C/B) / (p(A).p(C/A) + p(B).p(C/B))

### Law of total probability

Before discussing Bayes theorem we have to discuss law of total probability.

E1,E2 ..... En are n mutually exclusive and exhaustive events associated with a random experiment and an arbitrary event a is occurs with E1 or E2 or ... or En then

**Example:**

There are 5 boys and 6 girls in class A and 3 boys and 4 girls in class B. Two students are shifted from class A to class B. Then selecting two students from class B, what is the probability that selecting one boy and one girl?

**Solution:**

There are three probabilities for the shifting of two students from class A to class B

- shifting two boys
- shifting two girls
- shifting one boy and one girl

Let E1 be the event of shifting two boys from class A to class B.

E2 be the event of shifting two girls from class A to class B.

E3 be the event of shifting one boy and one girl from class A to class B.

N be the event of shifting one boy and one girl from class A to class B.

p(E1) = 5C_{2} / 11C_{2} = 2/11

p(E2) = 6C_{2} / 11C_{2} = 3/11

p(E3) = 5C_{1} * 6C_{1} / 11C_{2} = 6/11

If the event E1 has already occurred. i.e. two boys are shifted from class A to class B. Then the number of boys and girls in class B are 5 and 4 respectively.

p(N/E1) = probability of selecting one boy and one girl from class B after the occurrence of event E1.

= 5C_{1} * 4C_{1} / 9C_{2} = 5/9

Similarly,

p(N/E2) = 3C_{1} * 6C_{1} / 9C_{2} = 1/2

p(N/E3) = 4C_{1} * 5C_{1} / 9C_{2} = 5/2

The total probability

p(N) = p(E1).p(N/E1) + p(E2).p(N/E2) + p(E3).p(N/E3)

= [2/11 * 5/9] + [3/11 * 1/2] + [6/11 * 5/9]

= 10/99 + 3/22 + 30/99

= 40 / 99 + 3 / 22

= (80 + 27) / 198

= (107) / 198

### Probability : Expected value

In a random experiment, if each outcome is associated with a value x, then the expected value of the experiment is

There for, Σ Pi is the sum of probabilities of the distinct events associated with the given random experiment.

Expected value = Σ PiXi

**Example**

A game coin throwing is based on an arrangement of 9 boxes arranged in a 3 * 3 square of different boxes form as in the given diagram. Each box addressed with different colours. There are 4 boxes in red color, 3 boxes in green color and remaining two boxes are in yellow color.

R | G | R |

Y | R | G |

G | Y | R |

When the participant is throwing a coin, if the coin is fall into any of the green box, the participant will get Rs.10.If the coin is fell into an yellow box he won't get or loss any amount. If the coin fell into a red box, the participants will loss Rs.7. What is the expected value of the game?

**Solution:**

Let 'R' be the event of the coin fell into a red box.

'G' be the event of the coin fell into a green box.

and 'Y' be the event of the coin fell into an yellow box.

P(R) = 4/9

P(G) = 3/9

P(Y) = 2/9

Expected value = (P(G) * 10) + (P(Y) * 0) + (P(R) * (-7))

= 3/9 * 10 - 4/9 * 7

= 30/9 - 28/9

= 2/9 tends to 0.22