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Probability for Non-Mutually Exclusive Events

Why we have separate formulas for Mutually Exclusive and Non-Mutually Exclusive Events?

Gagandeep Singh
3 min readFeb 1, 2022

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I always hated probability because I couldn’t understand it. I knew which formula to use but couldn’t understand why.

In this article I will try to simplify some of these concepts.

What are Mutually Exclusive Events?

When you flip a coin there are always 2 outcomes, it can be either a heads or a tails. They are mutually exclusive because only one of them can happen at a time.

p(heads) = 1/2

Similarly, if you roll a dice, you’ll always get a single number. It can range from 1 to 6. This is also an example of Mutually Exclusive Event.

p(getting 6 when a dice is rolled) = 1/6

What are Non-Mutually Exclusive Events?

Let’s see you are watching the news and there is a 10% chance of rain the next day. You also know that tomorrow is your work day off and you have to wash your clothes. So, you decide to wash the clothes anyways whether rain happens or not. They are non-mutually exclusive events because the outcome of other doesn’t affect the other.

Now, we have simplified the concepts, let understand how to calculate probability for these events.

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1. What will be the probability of getting a number greater than 5when a dice is rolled?

So, we have to find the probability of getting a number greater than 5.

We only have 6 which is greater than 5. So, the probability will become —

P(getting number greater than 5) = 1 / 6

2. What is the probability of getting heads or a number greater than 5?

When we are talking about OR in probability it means ADDITION. This statement hold true for mutually exclusive events but fails for non-mutually exclusive events. Let’s see why

Let’s say we want to find the probability of getting heads or tails when a coil is tossed.

p(heads) = 1/2 and p(tails) = 1/2

so, p(heads) or p(tails) = 1/2 + 1/2 = 1

Now, if we apply similar concept for 2nd problem,

p(heads) = 1/2 and p(getting number greater than 5) = 5/6

p(heads) or p(getting number greater than 5) = 1/2 + 5/6 = 8/6

Something feels wrong here, how can the probability go above 1?

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The problem is that we are double counting. How? Let’s see

  1. Let’s say you flip a coil and roll a dice and expect heads
  • Heads — 1
  • Heads — 2
  • Heads — 3
  • Heads — 4
  • Heads — 5
  • Heads — 6

2. Now, lets say you flip a coin and roll a dice and expect a number greater than 6

  • Heads — 6
  • Tails — 6

P(heads) = 6/12 — we have listed 6 outcomes out of total 12 outcomes.

P(number greater than 6) = 2/12 — we have listed 2 outcomes out of total 12 outcomes.

There are a total of 8 outcomes but only 7 are unique and this double counting makes the result incorrect.

So, how do we correct it?

P(heads) or P(Num>6) = P(head) + P(Num>6) - P(heads, Num>6)

where P(heads, Num>6) = P(heads) X P(Num>6) = 1/2 X 1/6

We’ll have to subtract the duplicates.

I hope you we able to understand the concept. If you have any questions, please post them in the comment section.

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Gagandeep Singh
Gagandeep Singh

Written by Gagandeep Singh

Data Scientist | GenAI | NLP | Predictive Analytics | Speech Analytics

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