- Introduction
- Case1 : Probability of Particular Actor getting selected
- Case 2: Probability of One lady getting selected
- Case 3: One man and one woman (“AND” case)
- Case 4: “Atleast” one man is selected (“OR” case)
- Case 5: Probability of Not getting selected
- Mock Questions for CSAT, CMAT, IBPS, State PSC
- Recommended Booklist
- My Previous Articles on Aptitude

# Introduction

see url Probability is merely a subtype of Permutation and Combination (PnC) concept.

Probability can be solved without mugging up formulas, if your basic concepts are clear.

So, before proceeding in this article, DO read my previous article : Permutation & Combination (PnC) made easy without formulas

Since I’m not interested in talking about coins, dice and card in probability article, Let’s get back to Gokul-dhaam Society.

We’ll learn all the concepts of probability, with following 6 characters so keep the names and faces in mind.

# Case1 : Probability of Particular Actor getting selected

Director Asit Modi wants to shoot an episode in Essel World waterpark, he has to pick up only one character out of the given six character. What is the probability that Jethalal will be selected for this episode?

How many ways can you choose 1 character out of the given talent pool of 6 characters?

Fundamental counting principle: 6 ways. (or 6C1 =6)

Means total possible results=6.

And what is asked in the question?

Jethalal should get selected.

So keep Jethalal aside in a separate ‘talent-pool’ and ask your self, how many ways can Jethalal be selected out of given talent pool?

There is one character and we’ve to pick up one character: 1C1= One way only.

Desired result = 1 ways

Matter is finished.

Probability = Desired result divided by total results = 1/6

See the photo, it should make the picture clear

Now I’m modifying the case

# Case 2: Probability of One lady getting selected

Quest. Same question as before. Modi has to pick up one character out of the given six. What is the probability of a lady getting selected?..Continue Reading

Total results remains the same: 6C1= 6 ; as in case #1

And what is asked?

One lady.

How many ladies do we have in this talent pool? 2 (Anjali and Babita)

Put them in separate ‘ladies’ talent pool.

How many ways can you select one lady out of the given two ladies? : 2 ways (2C1) = desired result

Probability = 2/6 = 1/3

## Alternative: Same question rephrased:

What is the probability that the selected one character has moustaches?

Out of the given six, only two has moustaches (Jethalal, Mehta-saab)

Then principle remains the same as in Case #2

# Case #3: One man and one woman

Q. Modi has to select a team of 2 characters out of given six. What is the probability that it has one man and one woman?

Again this is a combination problem, we’ve to select a ‘committee’ of two people. Jetha-Mehta (JM) is same as Mehta -Jetha (MJ)

## Total results

How many ways can you select two characters out of given six?

= (6 x 5) / (2 x 1) =15 (=6C2)

How did we get this ^?

Your “combination” concept should be rock-solid now. If not, read my previous PnC article again.

## Desired result

Selection of one man and one woman.

Recall the very first example on “fundamental counting”.

Jethalal had to pick up two trains. (Mumbai – A’bad and Ahmedabad- Kutch). Same principle is applied here

=[pick one man out of given 4] (multiply) [pick one woman out of given 2]
=4 x 2

=8 ways.

When “** prednisone 500 mg AND**” is asked. You ** http://cinziamazzamakeup.com/?x=farmacia-viagra-generico-a-Napoli multiply** the cases.

You can also look at this as a combination problem

Choose one man ‘Committee’ out of 4 men and choose one woman Committee out of 2 women.

=4C1 x 2C1

=4 x 2 = 8 ways.

Desired result = 8 ways.

Final answer

Probability = Desired result / Total results = 8/15

# Case 4: “Atleast” one man is selected

Ques. Direct Asit Modi has to select 2 characters out of given six. What is the probability at atleast one of them is a man.

When they ask “Atleast or Atmost” etc we’ve to break up the case.

Team of two actors can be formed in following fashion

1. 1 man and 1 woman. (Order doesn’t matter, this is a Committee!)

2. Both men

3. Both women

Atleast 1 man in a team of two memebers means

= [1 man and 1 woman] OR [2 men]
Atmost 1 man in a team of two members

=[Zero man, 2 women] or [1 man, 1 woman]
When “OR” comes, we’ve to ADD (+) the probabilities.

Total results Is same : 6C2= 15 ways.

Desired results = [1 man 1 woman] (OR)+[ Two men]

## Situation #1: One man and One woman

Desired result =8 ways. ( we already did this in Case #3)

## Situation #2: Two men

Concept is same as Case #2

There are four men: Jethalal, Mehta-saab, Bhide, Aiyyar

Keep them aside in a seperate ‘talent-pool’

How many ways can you pick up 2 actors out of the given 4?

Combination problem

= (4 x 3) / 2 x 1

=6 ways.

## Final answer

Probability = Desired result divided by Total results

= (Situation #1 OR situation #2) / total result

= (8+6) / 15 ; because OR means addition (+)

=14/15

# Alternative method: 1 Minus

Total sum of all probabilities =1 (according to theory)

We had to select 2 actors out of given 6. So gender wise three situations possible in this ‘team’

1. Two men (MM)

2. Two women (FF)

3. One man and One Woman (MF)

Total probability (1)= MM + FM + MF

Our question was Atleast one man hence (MM+FM)

So, Probability of atleast one man

=1 minus probability of two women (FF)

MM +FM =1 – FF

So find the probability to two women getting selected?

Desired result= 2 women

Same as case#2

Keep Anjali and Babita in a separate talent-pool.

How many ways can you select two actresses out of the given two actresses?

Obviously 1 way. (2C2); because you’ll have to take both of them!

Total result= already counted 6C2= 15

Probability (FF) = 1/15

So our question

= 1 – (1/15)

=14/15

Compare the answer. IT is same what we got in the first method of Case#4.

# Case 5: Not getting selected

This one is very easy but writing it just in case Alternative method of case #4 did not click your mind.

Ques. Direct Asit Modi wants to select one character out of the given six. What is the probability that Anjali will not be selected?

Method 1: Direct method

Total results= 6C1= 6 ways (to select 1 character out of given 6)

Desired results

We don’t want Anjali in the selection so we keep Anjali aside and create a new talent pool of five remaining characters

(Jethalal, Mehta-saab, Bhide, Aiyyar and Babita-ji)

Pick any one out of this five and we’re done, we’ll satisfy the desired result that Anjali is not selected.

So, How many can you pick up one character out of five? = 5 ways (5C1)

Hence

Final probability = desired results divided by total results

=5/6

# Indirect Method: One minus other probabilities

According to theory, Total sum of all probabilities =1

So Probability of Anjali NOT getting selected

= 1 minus [Probability of Anjali getting selected]
Just like case #1, Probability of Anjali getting selected is 1/6

=1 – 1/6

=5/6

^Formula: P(E)=1-P(E’)

# Mock Aptitude Questions from Probability topic

In your UPSC CSAT, CMAT, State PSC or Bank PO (IBPS) exams, most probably they’ll ask you very cliched questions about coins and dice, bags and balls, or card games.

Let’s solve a few.

# Dice Probability question

Q. Two dice are tossed once. What’s the probability that both of them will show even numbers?

Concentrate on one dice only.

A dice has 6 numbers on it : 1,2,3,4,5,6

You toss it once, and one number will appear.

Total results = 6C1= 6 ways.

Desired result: We want even number.

Even Numbers are three: 2,4,6

How many ways can one Even number appear out of the given three? 3C1=3

Probability of one dice showing even number

=Desired result / Total results

= 3C1/6C1

=3/6

=1/2 { Fixed a silly mistake here, Thanks Mr.Manikant for pointing it out. }

Probability of two dice showing even numbers

= first dice shows even number (AND) second dice also shows even number

=1/2 x 1/2 ; AND means multiplication

=1/4 ; final answer

# Bags and balls: Replacement killers

Q. A bag contains four black and five red balls, if three balls are picked at random one after another without replacement, what is the chance that they’re all black?

Total balls = four black + five red = 9 balls

## Shorter Method

** buy branded cialis Total Results:** How many ways can you pick 3 balls out of 9 balls?:

Come on this is just another “Combination” problem, 9C3= (9 x 8 x 7) / (3 x 2 x1) =84.

** miglior sito per comprare viagra generico 25 mg spedizione veloce a Parma Desired Result:** How many ways can you pick 3 balls out of given 4 black balls?: 4C3 =4 ; (choosing 3 out of given 4, is same as rejecting one out of given 4 means you can do it 4 ways)

Final Probability

=Desired / Total

=4C3/9C3

=4/84

=1/21

## Longer Method and Explanation for this Bags and Balls problem

We have to pick up three balls one after another and we want them to be black. Means

1st picking black AND 2nd picking is black and 3rd picking is black; “And” means multiply.

It specifically mentions that “without replacement”, meaning everytime balls will decrease and our total events will change accordingly.

## 1st pick

Total results = pick one out of given 9= 9 ways

Desired= pick one out of given 4 black=4 ways

Hence, 1st picking probability =4/9

## 2nd pick

Now we’ve total 8 balls left, 3 of them are black

Run the same procedure of first pick: and you get probability 3/8

Same way third pick: 2/7

## Final probability

=first pick black AND x second pick black AND third pick black; (“And” means multiplication)

=(4/9) x (3/8) x (2/7)

=1/21

Now compare this with the shorter method explained previously

Essentially (4/9) x (3/8) x (2/7) is what you get when you expand the Formula: 4C3 / 9C3. (Because 3! factorial will cancel eachother)

# Balls WITH Replacement

Q. A bag contains four black and five red balls, if three balls are picked at random one after another WITH replacement, what is the chance that they’re all black?

It is the same questions like previous one but, WITH replacement means you pick up a ball note down its color and then put it back in the bag again. So Total Number of balls remain same for each event. And Hence probability of picking a black ball (4/9) remains the same in every case.

1st Pick up : 4/9 ; as calculated above under “1st Pick” topic

2nd Pick up: 4/9 ;because we put the ball back in the bag, So probability is same as “1st Pick”.

3rd Pick up: 4/9 ;because we put the ball back in the bag.

So final probability

=1st x 2nd x 3rd

=4/9 x 4/9 x 4/9

=Cube of 4/9

=64/729

## The End Not Yet

This is merely a tip of the iceberg. To get a firm grip over Probability topic, You must practice as many sums as you can. Otherwise you’ll make silly mistakes in calculation or get stuck for 7-10 minutes in a simple sum inside the exam hall. Practice is the KEY to attain perfection.So open your books, start solving.

## 23 Comments on “[Aptitude] Probability Made Easy (Extension of Permutation Combination Concept!)”

the right hand side of page not viewable….yhanx

Nice post…

Please post more such articles…

it is very much intresting method of learning math

In case 4 answer should be 4/5.

sir amazin effort….god bless you..

sir Please help

A bag contains four black and five red balls, if three balls are picked at random one after another without replacement, what is the chance that they’re all black?

The same question with replacement ..

How can it be done with Shorter Method given above

I tried it but couldntget the same answer as 64/729 given above

Hi,

For Case 4.

Can we try to solve it like Dice Probability question.

Prob of getting even Number in first dice is 1/2

Similarly Prob of selecting male for first chair is 4/6

Prob of selecting even number in second dice is 1/2

Prob of selecting male/female for second chair 5/5=1

Total Prob = (4/6)*1 = 4/6 = 2/3.

Using the alternative method mentioned in case 4 I can understand that the concept I mentioned is wrong. But I cannot find what is wrong with this approach.

I never checked this portion, I never new that you write on these topics too… you are all rounder – Mrunal

for case 3 HW did 15 total came???

sir, thank you so much. i do somewhat perfect in other topics of apti. but i was always struggling with probability and permutations and combinations. now m clear of concepts. pls continue ur guidance for people like me.

sir, i m confused totally in solving one question… pls help me.

question: find the number of combinations that can be formed with 5 oranges,4 mangoes and 3 bananas when it is essential to take atleast one fruit.??? answer is 119… pls help me sir…

sir, this question was solved by one formula.

if out of(p+q+r+t)things, p are alike of one kind, q are alike of second kind and r are alike of third kind and t are different, then the total number of combinations of one or more things is equal to

(p+1)(q+1)(r+1)2^t-1

this s it…

answer is given as: the required number of combinations

=(5+1)(4+1)(3+1)2^0-1

=120-1

=119

i couldnt understand… how can i do it without this formula??? pls sir help me with this… i tried many possible ways, but each time ended with confusion…

Sir Please provide guidance for NICL forthcoming exam will held on sept 8.

hi mrunal, one simple doubt.

Question : 2 cards drawn frm pack of 52 cards, prob. of getting both black or both queens?

in the above question , there will be 3 conditions

a) gettting both black

b) getting both queen

c) getting queens of black card.

for count cond a), we can say its (52X51)/2

for count cond b), we can say its (4X3)/2

for count cond c) , how do we do it ??

how do i caculate for the cond c) ??

is their -ve marking for acio exam ,

Question : 2 cards drawn frm pack of 52 cards, prob. of getting both black or both queens?

in the above question , there will be 3 conditions

a) gettting both black

b) getting both queen

c) getting queens of black card.

for count cond a), we can say its (52X51)/2

for count cond b), we can say its (4X3)/2

for count cond c) , how do we do it ??

how do i caculate for the cond c) ??

answer:

for cond 1: both cards to be black, there are 26 black cards in a deck so selecting 2 black cards in 26C2 ways.

for cond 12: both cards to be queen, there are 4 queens in a deck so selecting 2 queen in 4C2 ways.

for cond 3: both cards to be black and queen, there are only 2 such queens in a deck so selecting 2 cards in 2C2 ways.

total ways of selecting any 2 cards is 52C2.

so prob: (26C2 + 4C2 +2C2)/52C2.

Two dice are tossed. The probability that the total score is a prime number is:

A. 1/6

B. 5/12

C. 1/2

D. 7/9

Sir how to proceed with this dice problem..

15/36 i.e. 5/12 is the answer

Prime Nos: 2,3,5,7,11

Desirable Combinations: (1,1) (1,2) (2,1) (1,4) (4,1) (2,3) (3,2) (1,6) (6,1) (2,5) (5,2) (3,4) (4,3) (5,6) (6,5) : Total 15 such combinations

All Possible Combinations: 6 X 6 = 36

Required Probability = 15/36 = 5/12

A cricketer can make century with 4 and 6 only in how many ways, pls explain how to splve

4^x(6^y)=100… so we knw dt 4 ×3=6×2 meanz we cn exchange 3 4’s for 2 6’s.. then initially 4^25 × 6^0=100…similarily 4^22× 6^2=100….. 9 times

In how many ways 10 mangoes can be divided among 4 children?

Precise and crystal clear explanations!!

Thanks for this effort!!