# Key Lessons from Probability LC-01

### Sample space:

It is the set of all the possible outcomes of an experiment. \*Experiment being anything from rolling a dice to tossing a coin\*.

### Event:

An event is a subset of the sample space.

### Naive definition of Probability:

P(E)= no. of favorable outcomes/ no. of possible outcomes.

Assumptions made:

* All outcomes are equally likely(it is a fair coin).
    
* We have a finite no. of sample space.(else the denom. will be infinity)
    

## Counting rules:

### Multiplication rule:

If you have an experiment 1 with n1 possible outcomes, and another experiment 2 with n2 possible outcomes for each and every outcome of experiment 1 and similarly experiment r with nr possible outcomes for each and every outcome of experiment r-1 then,

Total overall possible outcomes = n1 x n2 x … x nr. (i.e. , the product of all the possible outcomes).

example: imagine you go to eat ice-cream at a parlor and you have two cone options and 3 flavor options,

experiment 1: choosing either one cone from the 2 options. That gives us the no. of outcomes of exp 1 as 2.

experiment 2: choosing either one flavor from the 3 options. That gives us the no. of outcomes of exp 2 as 3.

so the total no. of outcomes or ways u can have a ice cream is = 2.3 = 3.2 = 6.

P.S. you can also imagine it as a tree with branches coming out to understand it more structurally.

### Binomial coefficient:

nCk = n!/(n-k)!.k!

#Subsets of size k, of group of n people where order doesn’t matter.

we can explain the above the expression with the following example:

Suppose you have **n people** (say, 5 people: Alice, Bob, Charlie, David, and Eva), and you want to **choose k of them** (say, 3 people) to go on a trip.

Let’s walk through what’s happening when we calculate:

nCk = n.(n-1).(n-2)….(n-k+1)/k!

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### 🎯 Step-by-step with an example:

Let’s take **n = 5**, **k = 3** (you want to choose 3 out of 5 people).

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### 🔢 First: Count **ordered selections**

If you select **3 people** *in order*, like:

* First person
    
* Second person
    
* Third person
    

You could pick:

* Alice → Bob → Charlie
    
* Bob → Charlie → Alice
    
* Charlie → Alice → Bob  
    ... and so on.
    

This counts **permutations**, which is:

5⋅4⋅3=60

Because:

* First choice: 5 options
    
* Second choice: 4 left
    
* Third choice: 3 left
    

This counts **all different ways to pick 3 people in order**.

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### ❓But wait — what if we **don’t care about order**?

Like if Alice, Bob, and Charlie go on a trip, it doesn’t matter if we say:

* Alice → Bob → Charlie
    
* Charlie → Bob → Alice
    
* Bob → Alice → Charlie  
    …it’s the same group.
    

So for **every group of 3 people**, there are **3! = 6 ways to order them**.

That means we **overcounted** by a factor of 3!.

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### ✅ So we correct for that:

5C3 = 5.4.3/3! = 10

There are **10 unique groups of 3 people**, ignoring the order.

## Question based on learnings:

### What is the probability of getting a full-house in poker?

P(f.h) = 13.(4C3).12.(4C2)/(52C5)