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)