If GMAT probability questions are giving you a hard time this blog might just be all the help you need to get started. In this blog, we will learn how to approach and solve GMAT probability questions in the simplest way possible. So shall we?

#### IN THIS BLOG:

1. GMAT Probability Concepts

2. The use of conjunction AND

3. The use of conjunction OR

4. Combination of ‘AND’ and ‘OR’

## GMAT Probability Concepts

As we now move towards the mathematical aspects of the topic, one underlying factor that recurs in every question of probability is that whenever one is asked the question, what is the probability? The immediate question that arises or should arise in one’s mind is the probability of what?

The answer to this question is the probability of the EVENT. The **EVENT** is the cornerstone or the bottom line of probability. Hence, the first objective while trying to solve any question in probability is to **define the event**. The event whose probability is to be found out is described in the question and the task of the student in trying to solve the problem is to define it.

In general, the student can either define the event narrowly or broadly. Narrow definitions of events are the building blocks of any probability problem and whenever there is a doubt about a problem, you are advised to get into the narrowest form of the event definition. The *difference *between the **narrow and broad definition ****of the event** can be explained through an example:

*GMAT Probability Questions:*

- What is the probability of getting a number greater than 2, in a throw of a normal unbiased dice having 6 faces?

The broad definition of the event here is getting a number greater than 2 and this probability is given by 4/6. However, this event can also be broken down into its more basic definitions as:

The event is defined as getting 3 or 4 or 5 or 6. The individual probabilities of each of these are 1/6, 1/6, 1/6, and 1/6 respectively.

Hence, the required probability is 1/6 + 1/6 + 1/6 + 1/6 = 4/6 = 2/3.

Although in this example it seems highly trivial, the narrow event-definition approach is very effective in solving difficult problems on probability.

In general, event definition means breaking up the event into the most basic building blocks, which have to be connected together through the two English conjunctions— ‘**AND’** and** ‘OR’**.

## GMAT Probability Concepts: The Use of the Conjunction ‘AND’

Whenever we use AND as the natural conjunction joining two separate parts of the event definition, we replace the AND with the multiplication sign. Thus, if A AND B have to occur, and if the probability of their occurrence is P(A) and P(B) respectively, then the probability that A AND B occur is got by connecting P(A) AND P(B). Replacing the AND by multiplication sign, we get the required probability as P(A) X P(B)

*GMAT Probability Questions:*

- If we have the probability of
*A*hitting a target as 1/3 and that of*B*hitting the target as 1/2, then the probability that both hit the target if one shot is taken by both of them is got by?

**Event Definition: ***A *hits the target AND *B *hits the target.

*P(A) *x *P(B) = *1/3 x 1/2 = 1/6

Note that since we use the conjunction AND in the definition of the event here, we multiply the individual probabilities that are connected through the conjunction AND.

## GMAT Probability Concepts: The Use of the Conjunction ‘OR’

Whenever we use OR as the natural conjunction joining two separate parts of the event definition, we replace the OR with the addition sign.

Thus, if *A *OR *B *has to occur, and if the probability of their occurrence is *P(A) *and *P(B) *respectively, then the probability that *A *OR *B *occurs is got by connecting *P(A) *OR *P(B). *Replacing the OR by addition sign, we get the required probability as *P(A) + P(B).*

*GMAT Probability Questions:*

- If we have the probability of
*A*winning a race as 1/3 and that of*B*winning the race as 1/2, then the probability that either*A*or*B*winning a race is got by?

**Event Definition: ***A *wins OR *B *wins.

** **P(A) + P(B) = 1/3 +1/2 = 5/6

Note that since we use the conjunction OR in the definition of the event here, we add the individual probabilities that are connected through the conjunction OR.

Have you taken the GMAT before?

## GMAT Probability Concepts: Combination of ‘AND’ and ‘OR’

If two dice are thrown, what is the chance that the sum of the numbers is not less than 10?

**Event Definition:** The sum of the numbers is not less than 10 if it is either 10 OR 11 OR 12. Which can be done by:

(6 AND 4) OR (4 AND 6) OR (5 AND 5) OR (6 AND 5) OR (5 AND 6) OR (6 AND 6)

that is, 1/6 x 1/6 + 1/6 x 1/6 + 1/6 x 1/6 + 1/6 x 1/6 + 1/6 x 1/6 + 1/6 x 1/6 = 6/36 = 1/6

The bottom line is that no matter how complicated the problem on probability is, it can be broken up into its narrower parts, which can be connected by ANDs and ORs to get the event definition.

Once the event is defined, the probability of each narrow event within the broad event is calculated and all the narrow events are connected by Multiplication (for OR) to get the final solution.

*GMAT Probability Questions:*

- In a four-game match between Kasparov and Anand, the probability that Anand wins a particular game is 2/5 and that of Kasparov winning a game is 3/5. Assuming that there is no probability of a draw in an individual game, what is the chance that the match is drawn (Score – 2-2).

For the match to be drawn, 2 games have to be won by each of the players. If ‘A’ represents the event that Anand won a game, the event definition for the match to end in a draw can be described as [The student is advised to look at the use of narrow event definition.]

(A&A&K&K) OR (A&K&A&K) OR (A&K&K&A) OR (K&K&A&A) OR (K&A&K&A) OR (K&A&A&K)

This further translates into: (2/5)^{2} (3/5)^{2} + (2/5)^{2}(3/5)^{2} + (2/5)^{2}(3/5)^{2} + (2/5)^{2 }(3/5)^{2 }+ (2/5)^{2}(3/5)^{2} + (2/5)^{2}(3/5)^{2} = (36/625) x 6 = 216/625

After a little bit of practice you can also think about this directly as:

^{4}C_{2 }x (2/5)^{2 }x ^{2}C_{2 }x (3/5)^{2 }= 6 x 1 x 36/625 = 216/625

Where, 4C_{2 }gives us the number of ways in which Anand can win two games and 2C_{2 }gives us the number of ways in which Kasparov can win the remaining 2 games (obviously, only one).

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