Probability comes for like 1 mark or 2 … the good is it is easy … the bad is it comes only for a mark or two!

On requests, I have decided to take the plunge and do a piece on probability – all dice, coins and cards come out and play!

So,

### 1. What is probability?

Probability is the__chance__of the happening or non-happening of

__event__; denoted by ‘P’.

### 2. What are events and sample space?

__Sample space__is the total number of occurrences that can happen.

__Event__is the occurrence of ‘something’ which we are concerned with.

For example: Jai and Veeru did that coin toss to gamble away their lives – awesome – but it has got an important lesson of probability too.

One coin – what are the possible out comes? – Two, as there can be a Head or a Tail.

Therefore, our

Say, Veeru wanted Heads –

Therefore, our

**sample space (S)**= 2 =**total number of possible outcomes**(*either head or tail)*.Say, Veeru wanted Heads –

**how many heads is possible in one coin?**– One. Thus 1 is our**Event (E)!**

### 3. How to find Probability

**is the chance of the occurrence of an event;**

__Probability__**[P = E/S]**

Thus, with E = 1, and S = 2,

Probability of Veeru going to die = E/S = ½ = 0.5! [But it was different in the film - I know, I know!]

**4.**Hold on now! If the chances of Veeru’s death is 0.5; what could be the chance of Jai being the one dying?

Again, ½ =0.5! [Think the film makers calculated only this probability!]

### 5. The ‘Non – event’.

Every event has its corresponding**‘non-event’**; which can be

**denoted as E'**. If the ‘event’ is happening, then non-event will not happen and vice versa.

If Veeru is going to die (E), then Jai won’t (E'); if Jai’s (E) going to die then Veeru (E') won’t!

Thus,

**P(E) + P(E') = 1**

In words, probability of event and probability of a non-event add up to 1.

Therefore,

**1- P(E) = P(E'),**

1- P(E') = P(E).

1- P(E') = P(E).

### 6. AND ‘n’ OR:

First off – AND is multiplying.OR is for addition.

If a question is worded like this – ‘if the probability of A hitting the target is 1/3 and B hitting the target is ½, what is the probability of

**A and B**,

**both**, hitting the target if a shot is taken by both.’

which means, P(A)

**AND**P(B) = P(A

**and**B hitting the target);

**P(A) x P(B)**

P(A) x P(B) = 1/6.

Now, if the question was worded - ‘if the probability of A hitting the target is 1/3 and B hitting the target is ½, what is the probability of

**A or B**hitting the target?’

which means, P(A)

**OR**P(B) = P(A

**or**B hitting the target);

**P(A)**+

**P(B)**

P(A) + P(B) = 5/6.

### 7. Some common sample space(s)!

For Coins | One Coin | Two Coins | Three Coins |

Sample Space (S) = | 2 | 2 x 2 = 4 | 2 x 2 x 2 = 8 and so on… |

For Dice | One Dice | Two Die | Three Die |

(S) = | 6 | 6 x 6 = 36 | 6 x 6 x 6 = 216 and so on… |

For Cards | Cards in one suit (Either Spade, Clubs, Hearts or Diamonds) | One Pack of Cards/ Deck = Total number of cards | Face Cards (King, Queen, Jack and Ace) of all the fours suits together |

(S) | 13 | 13 x 4 = 52 | 4 x 4 = 16 |

### 8. Concept of Odds:

Sometimes probability is viewed in terms of ‘odds for’ or ‘odds against’ an event.**= P(E)/P(E')**

*Odds in favour of an event***= P(E')/P(E)**

*Odds against an event,*

or,

Odds in favour of the non-eventor,

Odds in favour of the non-event

fairly simple, right? All you got to do is calculate the P(E) and the P(E'); then use the above formulae,

**if and only if the word ‘odds’**is in the question! Otherwise we calculate the normal probabilities as asked in the question.

That is all for today folks!

Hope this helps!

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