Verbal Prep I

Hey guys,

It has been a busy week for me and hence you have not seen many problems from my side. And I think I have covered most basics on probability. Every other problem would more or less be an extension of these. Of course I have not dealt heavily with Permutation and Combination in these problems since I think it should be learnt as a seperate set.

So here for the weekend is something light. 5 words that are good to learn but are usually not found in the verbal wordlists of CAT preparation material. These are supposed to be an addendum to your learning.

As I have always said, the best way to prepare for English is to read a lot.
Do read up the fundaes on how to crack English if you already havent - Click here

So here are the words

1. Speakeasy
2. Aperitif
3. Tempest
4. Canyon
5. Holler

Answer to Hit or Get Hit

Click here for the question

Some basic assumptions on the rational strategy
1. All of them dont choose to deliberately miss everytime :) Else the game is not on and the probability of each surviving is 1.
2. Ugly and Bad will not shoot at each other and shoot at Good when he is alive. Because if they kill each other, then Good will kill them in the next shot
3. Good will shoot Bad when both Ugly and Bad are surviving. He wants to kill the one with higher shooting capability to increase his chances of survival

One wayy to attack this problem is conditional probability.
ugly shoots at Good and there are two conditions - Good survives or he gets killed.
Then under each scenario, there are more conditions and you go on and on and finish the problem. As you have already guessed that is the most difficult way.

Let us take an relatively easier route cutting down conditions as much as possible.
1. The first decision is to be made by Ugly - Shud he shoot or deliberately miss. We have already determined that if he shoots, he shud shoot at Good.
A) If he shoots Good and hits successfully, then both him and Bad remain in the contest with Bad getting the first shot.
Then the probability of Ugly's survival = p(Bad misses)*p(Ugly wins starting first) = 1/2*1/2 (see below for calc)
{ p(Ugly wins starting first) = p(Ugly hits bad) + p(Ugly misses)*p(Bad misses)*p(Ugly wins starting first) = 1/3 + 2/3*1/2*p(Ugly wins starting first)
Hence p(Ugly wins starting first) = 1/2 }
B) If he deliberately misses Good, then the chances of survival od Good, Bad and Ugly are as follows
p(Good survival) = p(Bad misses Good)*p(Good hits Bad)*p(Ugly misses Good) = 1/2*1*2/3 = 1/3
p(Bad survival) = p(Bad Hits Good)*p(Bad wins duel with Ugly, with Ugly starting first) = 1/2*1/2 = 1/4
p(Ugly survival) = 1-1/3-1/4 = 5/12

Since 5/12 is greater than 1/4, Ugly would choose to go with the strategy of deliberately missing the first shot. Once A misses the first shot, the probability of each survival are as given in option B
Chances of Good survival = 1/3
Chances of Bad survival = 1/4
Chances of Ugly survival = 5/12

The trick in this question is that the best has the least chance of survival. Note that the probabilities would have been the same if Bad had started. But the probabilities would change if Good had started.

Hope you enjoyed wracking the brains

Problem 7 - Hit or get hit

Hey guys,

CAT Math is tricky and evolves every year. So here is an interesting problem for the weekend that will test your rational thinking apart from the basics of probability.

I am sure that most of you have see one of the all time best movies, "The good, the bad and the ugly" starring Clint Eastwood, Lee Van Cleef and Elli Wallach and remember the last scene of the movie where they have a 3 men duel.

Three people Good, Bad, and Ugly are playing a 'shooting at each other' game. Shooting attempts rotate cyclically Ugly first, Bad second, Good next and the order repeats. Ugly has 1/3 chance of hitting his target, Bad has 1/2 and Good has 1. A hit means that the target person gets killed and the person hit drops out.
What is the probability of survival for each of them under a rational strategy? They are also allowed to deliberately miss.

Note: Think of what is a rational strategy first

Click here for the answer

How to track this blog

Hi,

With many people visiting the blog, I thought that I shud post this old post of mine once again so that people can track what is happening on the blog in better ways than coming on the blog often and checking for updates

1. Join the Google CatFundae egroup at http://groups.google.com/group/catfundae. By this you get emails on all the comments posted by the people on the blog. Gmail ID is not necessary.

2. For those who are a little technologically advanced, you can use Feedburner RSS feeds by clicking on the orange button on the sidebar here. RSS feeds give you the ability to track any blog/website. You also have an option of email in this too, though all the updates may come only at the end of the day.

3. You can also track the blog on Technorati by adding to the favourites with the green button on the sidebar. Technorati is a leader in blog tracking and also rates various blogs.

4. Go through the links on the sidebar and it will link you through to all the portions of the blog. Problems, GDs, Knowledge sessions and fundaes on how to crack CAT

5. You can use the search on the sidebar if you know what you are looking for.

Happy surfing

Complete Math problems at one place

Here is a post that makes it easy for people coming in new to dig out all the math problems we have discussed till now.

For fundaes on how to crack Math problems in CAT, click here

Problems discussed till now
1. Fake Note problem
2. The Hidden balls
3. Guess and make a fortune
4. Boys and girls
5. Pick the higher number
6. Tennis ball or Cricket ball
7. Hit or Get hit

Answer to black or blue

The answer is 2/3.

To think simple, let us list out the equally likely scenarios.
1. First ball was tennis. Second ball was tennis. Rafael pulled first ball
2. First ball was tennis. Second ball was tennis. Rafael pulled second ball
3. First ball was cricket. Second ball was tennis. Rafael pulled second ball

Out of the three cases, two correspond to the first ball being tennis. Hence the answer is 2/3

Note: If we had not known that Rafael had pulled out Tennis ball, then there would be one more scenario - First ball was tennis. Second ball was tennis. Rafael pulled first ball. In that case the answer would have been 3/4

People go wrong when you take the conditional probability approach and fall into the 3/4 trap. You analyse the problem as "With two equally likely scenarios, what is the probability of getting a tennis ball". The answer for that question in 3/4. In this question, the difference is that the event has already happened and the question is the probability of the another ball also being a tennis ball.

Think deep and understand the difference I have stated above very carefully so that you dont get hunted by CAT.

And btw, the question name shud have been "yellow or red" :p

Click here for the question

Problem 6 - black or blue

Hope that the small things you are learning here are increasing your knowledge arrows in your quiver. CAT is near. Happy hunting.

Now the question
You have a ball in a bag. It is either a yellow tennis ball or a red cricket ball with equal likelihood. We dont know which one is inside.
Roger comes in and adds a yellow tennis ball into the bag.
Rafael comes and pulls out a ball from the bag. The ball he pulls out is a tennis ball.
What is the probability that the other ball remaining in the bag is a tennis ball.

The answers are 1) 1/3 2) 2/3 3) 1/4 4) 3/4

Hint: The probability was .5 before Roger added the ball. Since the four answers above are spread on both side of .5, first think if the probability shud increase or decrease.
Click here for the answer

Answer to pick the higher number

Click here for the question

All the people who answered hit it right.
My only input on this would be to arrive at the same solution in a simpler way with less calculation. As I have said in some of the previous problems, conditional probability could involve some calculations and in some cases this could be avoided. (remember the hidden balls problem).

So the solution:
1. Say Aish picks up a number X.
2. The probability of Abhi picking up the same X is 1/15
3. Apart from the above case, where it could be equal, the chances of Abhi picking a higher number is equal to the chances of Abhi picking a lower number.
4. Hence the probability = (1-1/15)/2 = 7/15

The learning from this is, when there are a set of cases which are mutually exclusive, equally likely and form a comprehensive set, there is a high probability that you have a simpler way of getting the answer than going through conditional probability.

Click here for the question

Problem 5 - Pick the higher number

There are 15 folded chits numbered 1 to 15. Aish comes and picks a chit, notes the number and puts it back into the lot. Abhishek comes and picks a chit from the lot. What is the probability that the chit Abhishek has picked up has number higher than the one picked by Aishwarya.

The rule as always is to 'think simple' :)

Answer to Boys and Girls problem

The answer is .4

This was a very simple problem. The key in probability is to keep your thinking as simple as the problem allows one to.

There are 16 permutations for the four children.
2*2*2*2 = 16.
GGGG, BBBB, BBGG, BGBG, BGGB, GGBB, GBGB, GBBG, GBBB, BGBB, BBGB, BBBG, GGGB, GGBG, GBGG, BGGG
Now, since it is already given that one of them is a girl, the permutation of BBBB is ruled out. That leaves 15 permutations.
In that count the ones that have just one girl. That would be six. BBGG, BGBG, BGGB, GGBB, GBGB, GBBG

So the probability is 6/15 = 0.4

Of course in CAT, you should arrive at 6 and 15 using permutation formula to make the calculation quicker

Problem 4 - Boys and Girls

What is the probability that a man having four children has two girls and two boys, given that he has at least one girl.

The answer choices are a) 0.375 b) 0.5 c) 0.4 d) None of these

Answers by tomorrow morning

Answer to Guess and make a fortune game

Click here for the question
The answer is that the contestant is better off making a switch.

It is very difficult to articulate the answer is such questions, but let me make a try. If people dont get it, then I will make more tries from different angles

Here goes the argument:
1. When the player chooses the box in the beginning of the game, the probability of him getting the fortune is 1/3.
2. Let us divide the boxes into two sets. One contains the box that the player has choosen and the second set contains the two boxes that the player has not choosen.
3. The probability that the fortune is in the second set is 2/3.
4. Now the host comes and removes an empty box from the second set. Now if the player is given a chance, he is better off choosing the second set because the probability of winning is 2/3

If you are still confused and half convinced, think that there were 100 boxes and fortune behind one. Now you choose one box and the host eliminates 98 empty boxes out of the rest 99. Wont you make a switch?

Click here for the question

Problem 3 - Guess and make a fortune game

In a game of 'guess and make fortune', the rules are as follows.
1. There are three boxes and hidden under one of them is a huge amount of money. There is nothing under the rest two boxes
2. The host knows where exactly the fortune is
3. The player has to choose one of the three boxes
4. After the player chooses a box, the host opens one of the remaining 'empty' boxes
5. The player is now allowed to switch his choice of box if he wishes

The question is, does the player have a higher probability of winning if he makes a switch.
As always, the rule to solve simple everyday probability problems is to think simple. Afterall CAT tests your thinking skills more than your ability to do complicated math

Click here for the answer