# CAT Quant: Solving Advanced Probability-Based Questions

*(Photo: Lydia)*

In my previous post on probability, we talked about the basics of the topic and I honestly believe that that much should be enough for most CAT applicants. But there is a finite probability that a slightly more advanced question on probability might be asked in CAT. The probability that a more advanced question on probability will be asked in XAT is a littler higher than the probability of it being asked in CAT. Confused? Don’t be – I am just being geeky here.

In the previous post, we had talked about **discrete variables** whereas today we will talk about **continuous variables**. The most common type of discrete variable is a count, in which case it must be a whole number. Examples include family size, number of road deaths and number of cigarettes smoked per day, result on throw of a dice. Continuous variables are defined on a continuous scale. Examples include length, weight, etc.

Let us look at some of the classic problems on probability with continuous variables and how to solve them.

**Broken Stick Problem**

* ***If you break a stick randomly in two places, what is the probability that you can form a triangle from the three pieces?**

To solve this, let us consider the length of the stick as 1 and the two points that it is broken at as ‘x’ and ‘y’ where x

So, the length of the three parts will be **x, y – x and 1 – y**

Let us represent this on a graph

The point P, which gives us the exact points at which the stick was broken will lie inside a square of side 1.

For the three parts to form a triangle, they should satisfy the triangle criteria that the sum of two sides is greater than the third side.

x + (y – x) > (1 – y)

=> y > ½

x + (1 – y) > (y – x)

=> 2y – 2x

=> y – x

(y – x) + (1 – y) > x

=> x

If we apply these conditions on the graph we will get,

To meet the criteria, P(x,y) has to lie in the triangle shown above.

The probability of that happening = Area of the triangle / total area = 1/8

The exact same pattern will be repeated when x > y, and we will get the probability again as 1/8 in that case.

**Required probability = 1/8 + 1/8 = ¼**

**Minimum Length Problem**

**Two points A and B are selected at random on a segment of length 2 cm. What is the probability that the length of the line AB > 1 ?**

* *Apply the exact same idea as discussed in the broken stick problem. Our basic graph would look like:

As you can seem the blue line is used to indicate that A = B.

We want the probability that the distance between A and B is greater than 1

=> The distance from the line A = B should be at least one unit

Using this restriction, we would get the graph as

So, our point P can lie in the either the triangle on the top-left or in the triangle on the bottom right.

Required probability = Sum of the areas of the triangle / Total area

= **(0.5 + 0.5)/4 = ¼ **

* *

** **

**Time Related Problem**

**Two ISB alumni decide to meet at Café Linger On between 9:30 am and 10:30 am. They agree that the person who arrives first at the cafe would wait for exactly 15 minutes for the other. If each of them arrives at a random time between 9:30 AM and 10:30 AM . What is the probability that the meeting takes place?**

* *

This can be solved exactly with the same concept of ‘Minimum Length Problem’.

The time period in which they can arrive is 1 hour and they have to be within 15 minutes of each other or 0.25 units of each other.

=> The distance from the line A = B should be less than 0.25 units

Required probability = 1 – Unfavorable probability

= 1 – (Sum of the areas of the two triangles / Total Area)

= **1 – 2*(½*¾*¾) = 7/16**

**Ravi Handa** has taught Quantitative Aptitude at various coaching institutes for seven years. An alumnus of IIT Kharagpur where he studied a dual-degree in computer science, he currently runs an **Online CAT Coaching and CAT Preparation Course** and the website **Handa Ka Funda.**