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With about a month or so to go, the question that junta is asking at this point is not "Do I have it in me to crack CAT?" as much as "Do I have it in me to crack me in crack CAT in a month?" Now let us presume that you present your problem t...

Ravi Handa
@ravihanda
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*Photo Credit: Steve A johnson*

Probability has always been a fascinating topic to teach. The pleasure that I get out of correcting silly mistakes, which are plenty in questions based on probability can be considered sadistic by some or a truly noble thing to do by others. I hope the percentage of the latter would be higher but I think the percentage of the former is higher. Let me try to redeem the reputation a little bit by talking about the basics of the topic, which most of the readers here might know but would be really important bit for those who don't.

**Funda #1**

**P(A) = Number of favorable outcomes / Number of total outcomes**

Probability of an event is defined by the above formula. Now where people make mistakes is in the understanding of the word 'outcome'. The 'outcomes', which are considered for probability, have to be equally likely in nature. They cannot be any random 'outcomes' of your choice. If you think this is obvious then let me give you an exercise:

What is the probability of getting a 4 on the throw of a regular dice?

Many of you would say the answer is 1/6 because there are six possible outcomes – getting a 1 or a 2 or a 3 or a 4 or a 5 or a 6; and out of these six possible outcomes one is favorable. If that is what you are thinking – you are right.

What do you think is wrong in my explanation if I say that the answer is ½ because there are two possible outcomes – getting a '4' and not getting a '4'; and out of these two possible outcomes one is favorable. What is wrong in this? I am covering all possible outcomes. The wrong part is that the 'outcomes' are not equally likely.

*In probability, the outcomes which are considered or counted have to be equally likely outcomes.*

**Funda #2**

**Exhaustive events** are those which will cover all the possible outcomes. For example, in the throw of a dice – getting a '1', getting a prime and getting a composite are exhaustive events because if you consider all of them together – they cover all the possible outcomes {1; Primes – 2,3,5; Composites – 4,6}

*For Exhaustive Events: P(A) + P(B) +P(C)… = 1*

** Mutually Exclusive events **are those which will have nothing in common between them. For example, in a race A winning the race and B winning the race are mutually exclusive events. It is possible that someone other than A or B might win the race but it is not possible that both of them will the race.

*For Mutually Exclusive Events: P(A ∩ B) = 0*

**Complementary events** are a sub-type of exhaustive events, which have nothing in common between them and there is no other event possible. For example, in a coin toss getting a head or getting a tail are complementary events. For example, in a class 'being a boy' or 'being a girl' are complementary events. For example, in case of shooting a target, hitting or missing are complementary events.

*For Complementary Events: P(A) + P(A') = 1 & P(A ∩ A') = 0*

*All Complementary Events are Exhaustive & Mutually Exclusive events.*

**Funda #3**

**Addition Rule:**

*P (A U B) = P(A) + P(B) – P(A **∩ **B)*

**Multiplication Rule**:

*P(A **∩ **B) = P(A) P(B/A) = P(B) P(A/B)*

For **Independent Events **P(A/B) = P(A) and P(B/A) = P(B)

ð*P(A **∩ **B) = P(A).P(B)*

ð*P (A U B) = P(A) + P(B) – P(A).P(B*)

**Funda #4**

The questions of the type when the same person tries to do something multiple times are pretty common in tests. A good example would be when you are trying to bomb a bridge in how many attempts the probability of the bridge getting blown would go above 80%. Or if you are an archer trying to hit a target, what is the probability that you will hit the target 4 times out of 7. These type of questions are primarily based on complementary events, so if the probability of one event is 'p'; then the probability of the other event will be '1 – p'

*If the probability of an event occurring is P, then the probability of that event occurring 'r' times in 'n' trials is = ^{n}C_{r} x P^{r} x (1-P)^{n-r}*

**Funda #5**

In cricket and more so in football, you might have seen the betting odds displayed on the top right / left of the sports page. Those odds are just a way of representing the probability of a particular team winning or losing. If the odds in favor of India in an India Vs. Australia match are 4:5; then the probability of India winning the match is not 4/5 but 4/9. Odds do not represent favorable and total outcomes but they give a ratio of the favorable and unfavorable outcomes.

**Odds in favor = Number of favorable outcomes / Number of un****favorable**** outcomes**

**Odds against = Number of unfavorable outcomes / Number of ****favorable**** outcomes**

I think this should suffice for an introduction to probability. I hope you liked this post and do provide feedback and / or ideas for future posts via the comment section.

**Ravi Handa** has taught Quantitative Aptitude at IMS for 6 years. An alumnus of IIT Kharagpur where he studied a dual-degree in computer science, he currently runs www.handakafunda.com

- 36 Comments
- Good article sir. Just a small correction-> For Independe.... 14 Mar '13.
- @ravihanda Sir, plz also discuss conditional probability.... 21 May '13.

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36Comments »Good article sir. Just a small correction-> For Independent Events P(A/B) = P(B) and P(B/A) = P(B) should be ->For Independent Events P(A/B) = P(A) and P(B/A) = P(B)

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