New Batches at TathaGat Delhi & Noida!
Probability
by Total Gadha - Sunday, 8 November 2009, 10:31 PM
  cat 2009 cat 2010 probability Dagny and I are big Harry Potter fans. I still read the books while eating or going to bed. At night when Dagny is screaming at me to turn off the lights and go to sleep, I have a sudden impulse to swish my ‘imaginary’ wand in air and yell ‘IMPEDIMENTA!’ Ah well, it is not advisable to throw spells at your wife, especially the one who is equally good at them, if you care about your marital bliss. So I resist the temptation. But the stranglehold of imagination is not lost on me. And imagination and dream building form a sizeable portion of our psyche if we are to get ahead in life. They also fuel our desire to focus and work hard towards our goals. I still remember that the only thing that propelled me hard towards cracking IIT JEE was that I wanted to win the heart of a girl in my school. I would dream about million fanciful situations in which she was looking at me wide-eyed with admiration after JEE results were announced. Alas, the dream did push me to my coveted institute but did not gave the courage to open my mouth in front of her. Even during our CAT preparations, Dagny and I both shared the common dream- that of staying together in IIM and traveling the world after passing out. Although I cracked CAT- twice- we realized that our dream was impossible while serving someone else. And then glory of an MBA fizzled out. We have let go of IIMs but we still cling on hard to our dream. Wish we have Floo powder or a Portkey. smile

This is a month when you should be stoking your dreams, trying to imagine what kind of life you want to lead, which places you want to go to, where do you want to live, what kind of home do you want, which girl do you want to propose to, etc… Take good care of your dreams. They will give you your lust for life, your raison de etre, your enthusiasm for CAT, and your killer instinct. If you want to really go for it, start with finding your dreams. smile

This lesson goes to Ashwini, whose continuous entreaties (TG sir please write a lesson on probability) in our CAT CBT Club forced me to sit up and finish the task. So if you want to say thanks for this lesson, say thanks to him. But yes, if you feel this lesson is useful to you, please do not forget to refer TotalGadha.com to your friends by clicking on tell and share feature below the login window. smile

probability 1

probability 2

probability 3

probability 4

probability 5

probability 6

 

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You might also like:
Permutation and Combination
Maxima, Minima and Inequalities

Re: Probability
by AsHwIn Drmz - Sunday, 8 November 2009, 10:53 PM
 

So finally ....A lesson on Probability.... smile

Thanks a lot Tg...

btw...m Ashwin not Ashwini smilesmilesmile...




Ashwin
Re: Probability
by amit kheterpal - Sunday, 8 November 2009, 11:10 PM
  thanks a lot tg sir and ashwin too.smile
Re: Probability
by wali khan - Sunday, 8 November 2009, 11:44 PM
  thanks a lot TG
smile
Re: Probability
by honey chaubey - Monday, 9 November 2009, 12:02 AM
  Thanks Sir
Re: Probability
by harsh tulsiyan - Monday, 9 November 2009, 12:56 AM
  thanks sir ..
it helped to clear many doubts ...

and thanks to ashwin..
Re: Probability
by steven hyde - Monday, 9 November 2009, 01:05 AM
  Sir,
One small doubt. In the beginning of this lesson when you mention the example of tossing 3 coins and say that that the possible outcomes are (HHH),....(TTT), isn't (HTH) equivalent to saying (THH) or (HHT) ? That is to say, are the 3 coins distinguishable? I sometimes take tossing 3 coins to be the same as tossing a coin thrice. Is there an error in this thinking?
Thanks .
Steven
Re: Probability
by KAMONASISH AAYUSH MAZUMDAR - Monday, 9 November 2009, 04:13 AM
  thank you so much :D for listening to our request always smile and so many other little things that we can't really thank you for.


now official answers to PnC lesson plz
Re: Probability
by whirl wind - Monday, 9 November 2009, 06:57 AM
  no steven,

When u take it as tossing 3 coins - u need to take care of which 2 coins are showing heads(or tails).(1st & 2nd coins or 2nd&3rd coins or 3rd and 1st coins).

If u take it as tossing the same coin thrice, u consider it to be the throw in which u get the heads.(on the 1st&2nd throws, or 2nd &3rd or 3rd&1st throws).

Nothing wrong in thinking tossing 3 coins to be the same as tossing the same coin thrice.

Whirlwind
Re: Probability
by shahnas ameer - Monday, 9 November 2009, 09:06 AM
  thankzzzzzzzzzzz
Re: Probability
by Ronak kabani - Monday, 9 November 2009, 10:49 AM
  Thnx TG
The probability of getting a sum correct on probability has definitely increased. smile

Ronak
Re: Probability
by shank jos - Monday, 9 November 2009, 11:29 AM
 

Thanks TG

Re: Probability
by Soumyakanta Das - Monday, 9 November 2009, 01:13 PM
  Thanks a Lot TG Sir and Ashwin... smile smile smile
Re: Probability
by divya kalkotwar - Monday, 9 November 2009, 01:48 PM
  thanks a lot TG smile
Re: Probability
by ashraf laskar - Monday, 9 November 2009, 02:11 PM
 

thanks a lot TG Sir for such a beautiful article .......

the funda to solve probability using geometry is new to me....hope that now i can attempt probability questions with more confidencesmile.......

-Ashraf

Re: Probability
by Vickram Asokan - Monday, 9 November 2009, 02:54 PM
  Hi TG,

Good article at a great time.

Have a doubt in the problems solved after the article.

Second from last question, which says,

" A,B and C are two speakers among 7 diff speakers "

Isn't that three speakers ( A,B and C ). Another thing is that there are two possible arrangements that are favorable for our criteria ["A" speaking before "B"]. ( i.e ABC and ACB are favorable).

Please correct me if I am wrong. Thanks for the article...

smile
-Vickram.
Re: Probability
by Priya Ranjan Dutta - Monday, 9 November 2009, 03:46 PM
 

Vickram,

Read the question carefully. A speaks before B who speaks before C.. only one case ABC..smile..

Re: Probability
by Neelesh Sethi - Monday, 9 November 2009, 08:10 PM
  Thanx a lot sir..probability by geometry were outstanding..sir how u thought of that method..it was difficult to even imagine that geometry can be used in probab..1 doubt ,If question is: if out of 6 speakers..find probability that abc speak after first speaker and in order, a followed by b followed by c? then what will be the answer
Re: Probability
by suraj verma - Monday, 9 November 2009, 08:48 PM
  smile superb TG sir...thanx a ton
Re: Probability
by rakesh salecha - Tuesday, 10 November 2009, 02:09 AM
  thank you sir
Re: Probability
by Suvrodip Banerje - Tuesday, 10 November 2009, 04:20 AM
 

Hello,

 Although am a keen reader of your articles,yet I have never posted any comment.But can't resist writing today,partly for the introduction part and partly to  point a very minor mistake(am sure it's a typo)..

In the question

In a match between two players A,B the chances of A winning are 3/5.If A and B play 5 matches in a succession,whta is the probability that B wins at least 3 matches(assuming no draws)?

the solution should be:

 5C3*(2/5)^3*(3/5)^2 + 5c4*(2/5)^4*(3/5) + 5c5*(2/5)^5.....

 

Dreamz Rule

Re: Probability
by Total Gadha - Tuesday, 10 November 2009, 10:55 AM
  Hi Surodip,

Corrected the typo. Thanks for pointing it out. smile

Total Gadha
Re: Probability
by Rohit Ghosh - Tuesday, 10 November 2009, 11:57 AM
 

Hi,

How the total probability of drawing a red ball is "19/45"?

Please help

Re: Probability
by abhay agrawal - Tuesday, 10 November 2009, 12:43 PM
  from time I've become a fan of your articles and always wait for the new one. I like one thing most in your articles and that is your explanation by examples. Although in two years of my preparation for CAT I never liked probability but after reading this article I'm really feeling very comfortable in it. And one thing more that after reading this article my probability in cracking CAT became higher.
So thanx to you and Ashwini for forcing you to write a wonderful article.
Re: Probability
by Total Gadha - Tuesday, 10 November 2009, 01:06 PM
  Thanks Lazy smile

Fill the 'Tell a Friend' to help us if you can?
Re: Probability
by Pradip Roy - Wednesday, 11 November 2009, 10:21 AM
 

Hi TG

Thanks a lot for the article.

I have a doubt about the Square-Coin sum.... can u explain that in a lil more detail  (centre of the coin is 1 unit inside of the square, so how did we get 8 units instead of 9 (10-1) units)

Re: Probability
by Rahul Arora - Wednesday, 11 November 2009, 11:05 AM
  Hi Pradip,

As the original square is of 10 unit length, so the inner square side = 10 - 1(from left side) - 1 (from right side), thus, 8. Hope, it helps you.
Re: Probability
by abhay agrawal - Wednesday, 11 November 2009, 12:22 PM
  Hi TG sir, I have question
Q.  There is a bag and in that bag there are two pockets. In one pocket of the bag there are 2 gold coins & 3 silver coins and in another pocket there are 3 gold coins & 2 silver coins. If a man choose one of pocket of the bag and pick up one coin from that pocket and place it in another pocket of the bag. Now he randomly pick up one coin from this pocket in which he place the coin than what is probability of getting a silver coin.
plz give the answer with the explanation
Re: Probability
by piyush jain - Wednesday, 11 November 2009, 08:03 PM
  well gr8 article with some wonderful probabilities smile
m big fan of urs but just one thing i cant make it out in prob. having 3 speakers A,B,C among 7 and v require to calculate probability of cases whr A speaks before B and B before C .

i tried it in this manner by fixing C at 1 position say last and now calculate as in case of A speaks before B among 6 speakers,again fix C at position 6 and calculate again same among 5 speakers similarly till C sit at 3 position .
so it comes out (360+60+12+3+1)/7!

so plz. explain this
piyush jain
plz help
by Avishek Chakraborty - Wednesday, 11 November 2009, 10:11 PM
 

guys I have one problem

numbers 1 to 20(1,2,3,...20) are arranged. wat is the probabilty that 2 will appear before any other even number?

Re: Probability
by Total Gadha - Wednesday, 11 November 2009, 11:44 PM
  Hi Lazy,

There are four cases here-
  • gold coin from Pocket 1 then silver coin from pocket 2- (2/5) ×
    (1/3)
  • silver coin from Pocket 1 then silver coin from pocket 2- (3/5) × (1/2)
  • gold coin from Pocket 2 then silver coin from pocket 1- (3/5) × (1/2)
  • silver coin from Pocket 1 then silver coin from pocket 1- (2/5) × (2/3)
Sum all these

Total Gadha


 

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Re: Probability
by Total Gadha - Wednesday, 11 November 2009, 11:50 PM
  Hi Piyush,

You are calculating it wrong in your method. For example, when A is in second position, you are doing that rest of the lower seats can be arranged in 5! ways out of which B would come before C in half. Unfortunately, you are not taking care of speaker no. 1 here whom you can select in 4C1 = 4 ways. So total number of ways are 4 × 60 = 240.

And so on. When A is in third place, you will have to select speaker no. 1 & 2 (not equal to B or C). Then you can do what you are doing.

Answer would be same by your method also. it would be 840/7!

Total Gadha


 

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problem
by Avishek Chakraborty - Wednesday, 11 November 2009, 11:55 PM
 

TG Sir since u r here thought if I can get the answer of this problemnumbers 1 to 20(1,2,3,...20) are arranged. wat is the probabilty that 2 will appear before any other even number?

thanks in advance

Re: Probability
by Avishek Chakraborty - Thursday, 12 November 2009, 12:26 AM
 

numbers 1 to 20(1,2,3,...20) are arranged. Wat is the probabilty that 2 will appear before any other even number?

I think the ans is 1/10.
See there are 10 even numbers in total. now there will be equal number of arrangements in which one even number comes first. like
numbr of cases in which 2 comes b4 ne othr evn no=number of cases in which 4 comes b4 ne othr evn number=...=number of cases where 20 comes b4 ne othr evn number
so the in this prob favourable cases=total cases/10=20!/10
so probability = 1/10
m I rite? can any1 confirm

Re: Probability
by abhay agrawal - Thursday, 12 November 2009, 10:56 AM
  hello sir, I'm taking this problem a little bit different and getting the same ans what u got. but official ans is diff. my take
if I choose any pocket for picking the coin then there will be only 2 possibility that it will be Gold or Silver. now
consider 1st pocket
I place gold coin in it than the probability of getting silver = 1/2 x 3/6
I place silver coin in it than 1/2 x 4/6
now consider 2nd pocket
I place gold coin than 1/2 x 2/6
I place silver coin than 1/2 x 3/6
Now sum all these and u will get the same ans. is my approach right?
Re: Probability
by Total Gadha - Thursday, 12 November 2009, 01:12 PM
  Hi Lazy,

My answer is coming to be 1/2 (you will have to multiply all the terms in my answer by 1/2 for choosing the first bag) but yours is different. Will think about your approach later. Please tell me the official answer for now.

Total Gadha
Re: Probability
by userdce . - Thursday, 12 November 2009, 06:34 PM
  content is great as ever but seems font got a bit distorted in the pictures
Re: Probability
by piyush jain - Thursday, 12 November 2009, 07:13 PM
  ok i got it sir thank u very much and congrats for 3rd anniversary of TG.
Re: Probability
by abhay agrawal - Friday, 13 November 2009, 11:34 AM
  yes sir, you r right. In my approach I was not considering the probability of picking a coin from any pocket. now i got my answer. thanx sir
Re: Probability
by ankhuri dubey - Friday, 13 November 2009, 08:34 PM
  Intelligent!!..smile i have a question Sir...wts d probability that the 'sarah n nidhi' in one of the questions weren't just a coincidence..??..:P
Re: Probability
by Total Gadha - Friday, 13 November 2009, 08:56 PM
  Ankhuri tongueout
Re: Probability
by Vickram Asokan - Monday, 16 November 2009, 12:42 PM
  Hi Ms.Priya,

Thanks for the clarification. I now understand why I am getting more negations in MockCATs... smile

Thanks,
-Vickram.
Re: Probability
by deepika gandhi - Monday, 16 November 2009, 02:27 PM
  pls...can u giv few more tips on functions,graphs and inequalities
Re: Probability
by Pankaj Kumar - Thursday, 19 November 2009, 12:19 AM
  Hi TG,

CAT is coming and by giving this article at this crucial time, u helped many as probability with permutations and combinations is bit challenging topic.
I know I'm becoming demanding here , but a humble and last request for this CAT- please give some tips for Logarithms topic also please..please please smile

I have immense appreciation for your great way helping many
TG will have to go much higher from here.., I encourage all people here to spread TG so that more n more people can benefit from it.

Many Thanks Again smile
Re: Probability
by Avishek Chakraborty - Thursday, 26 November 2009, 11:39 PM
 

guys,

7 distinct accidents occur in a week. what is the probability that they all occur on the same day of the week?? 

options are 1. (1/7)^7     2. (1/7)^6    3. (1/2)^7   4. 7/ 2^7

since since a single accident can occur on any single day so (1/7)^7 and for 7 days it becomes (1/7)^6. is it correct?

Re: Probability
by Abhijit Roy - Monday, 14 December 2009, 10:09 PM
  From very first time i Logged in to this site ...I find it very helpful ...n dis is nt only in regard to preparing my self for cat but also for other exam too.....

I was facing some problem wid probability ...n moreover being a non math student ..it was bit difficult for me to understand ......

So all thing is dat thank you very much for providing us such a useful n helpful site ......

It's a request to u sir if u can post something on Permutation n combination ... and Logarithm ....


Re: Probability
by gaurav adlakha - Tuesday, 26 January 2010, 04:20 PM
  great help.........
Re: Probability doubt
by vasu r - Monday, 25 October 2010, 10:52 PM
 

Paul the octopus who has been forecasting the outcome of FIFA world cup matches with tremendous accuracy has now been invited to predict ICC world cup matches in 2011. We will assume that the world cup contenders have been divided into 2 groups of 9 teams each. Each team in a group plays the other teams in the group. The top two teams from each group enter the semi finals ( after which the winner is decided by knockout).

 

        However, Paul has a soft spot for India and when India plays any team, Paul always backs India. Alas, his predictions on matches involving India are right only 2 out of 3 times. In order to qualify for the semi finals, it is sufficient for India to win 7 of its group matches. What is the probability that India will win the ICC world cup?

             a)    (2/3)^10                                  b) (2/3)^9 + 8/3 * (2/3)^9 

             c)    8/3 * (2/3)^9                           d) (2/3)^10 + 8/3*(2/3)^9

Re: Probability doubt
by kushal ganatra - Saturday, 30 October 2010, 11:49 AM
  i think answer should be D
Re: Probability doubt
by Pradeep Mavilla - Wednesday, 3 November 2010, 01:16 AM
  Thanks for the article
Re: Probability
by aagesh sharma - Monday, 27 June 2011, 06:26 PM
  applying 4 cbt is the irst thing i'm gonna do with this months salary thoughtful
Re: Probability
by manisha dalan - Wednesday, 3 August 2011, 02:50 PM
 

In the question-what is the probability that the red ball came from bag1...

is the Probability of drawing a red ball from bag1 not 3/5 instead of 1/5..

please correct me if m wrong.

thanks TG,

manisha

Re: Probability
by ADITYA PATEL - Wednesday, 10 August 2011, 11:36 AM
  Hello TG Sir,
Thanks for writing such a nice article.
I have 1 question.
in the example you gave for mutually exclusive events,(in rolling 2 dices, the probability of getting sum 4 or sum 7), you considered both the events as mutually exclusive, but i have a question why can't these events be considered as independent events and why can't we use probability equation P(A) * P(B)? I know sir i am wrong but i am not able to get the exact difference between Mutually exclusive events and independent events. Please do reply sir.
Re: Probability
by ritu aggarwal - Thursday, 11 August 2011, 04:48 PM
  sir..i have one ques of prob..

in poker, 5 cards are dealt to a person. assuming all C(52,5) are equallty likely. what is the probability of being dealt

(i) one pair, (a,a,b,c,d) where a,b,c,d are distinct?
(ii) two pairs (a,a,b,b,c) where a,b,c are distinct?

i didnt understand this question sir.
Re: Probability
by destiny unruled - Friday, 12 August 2011, 01:05 PM
  @ Aditya

Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring or vice-versa.

Now, if on the two dices sum of numbers showing up is 4, then it can definitely not be 7 which means they are not independent events.

On the other hand. two events are mutually exclusive if they can not occur at the same time, which is true in this case.

So, getting a sum 4 and a sum 7 are mutually exclusive events.
Re: Probability
by In lonely planet i live - Wednesday, 8 August 2012, 08:28 AM
  Well written article sir.

Please help me finding the answer for the below question.

A man alternately tosses a coin and throws a die beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is?

Thanks in advance.
Re: Probability
by TG Team - Wednesday, 8 August 2012, 10:47 AM
 

It is simply sum of infinite GP: 1/2 + (1/2)(4/6)(1/2) + (1/2)(4/6)(1/2)(4/6)(1/2) + ... = 1/2 + (1/2)(1/3) + (1/2)(1/3)(1/3) + ... = 3/4.

Re: Probability
by Rhythm Goyal - Wednesday, 8 August 2012, 07:46 PM
 

Hi TG,

First of all, Thanks a ton for such an nicely organised article. Often Probability confuse us and is a reason of dislike. Now we have a vision about how to approach a problem.

I have 3 querries related to your artical above and need your help for the same:


1) There is a question above as "10 persons are sitting around a round table. What is the probability that two particular persons are NOT sitting next to each other?"
I failed to undestand how we got 2*8!/9! as the  probability that 2 particular persons are sitting next to each other. Where did all this 8!, 9! and product by 2 came in?

2) There is a question in Inverse probability section as "Dagny's purse contains 4 coins which can be 1 rupee coins or 5 rupees coins. 2 coins are drawn and are both foubnd to be 5 rupees coins. If the coins are replaced in the purse and Dandny draws out another coin, what is the probability that it is a 1 rupee coin?"

As it was mentioned in the question itself that 2 coins of 5 are drawn, implied that alteast there are 2 coins of 5 rupee each in the purse.Thus there were 3 cases
a) all 5 rupee coin
b) 3 coins of 5 rupee and 1 coin of 1 rupee
c) 2 coins of 5 rupee and 2 coins of 1 rupee

Now why probability of drawing 2 5rupees coins were obtained?
also how come its 1, 1/2 and 1/6 respectively for case a, b,and c.
should it not be 1/3, 1/4, 1/6 respectively.

Further, I didnt understand how we have calculated the probablilty of case a, b & c.

3) In a question as "If we throw a dice 10 times what is the probability of it showing 6 exactly 3 times?"
Should not the solution be (1/6)^3 * (5/6)^7.
Why was it needed to found "ways of those 3 out of 10" i.e 10c3 and getting its product

 

Re: Probability
by Rhythm Goyal - Tuesday, 14 August 2012, 08:55 PM
 

Hi Kamal Sir,

First of all, Thanks a ton for such an nicely organised article. Often Probability confuse us and is a reason of dislike. Now we have a vision about how to approach a problem.

I have 3 querries related to your artical above and need your help for the same:


1) There is a question above as "10 persons are sitting around a round table. What is the probability that two particular persons are NOT sitting next to each other?"
I failed to undestand how we got 2*8!/9! as the  probability that 2 particular persons are sitting next to each other. Where did all this 8!, 9! and product by 2 came in?

2) There is a question in Inverse probability section as "Dagny's purse contains 4 coins which can be 1 rupee coins or 5 rupees coins. 2 coins are drawn and are both foubnd to be 5 rupees coins. If the coins are replaced in the purse and Dandny draws out another coin, what is the probability that it is a 1 rupee coin?"

As it was mentioned in the question itself that 2 coins of 5 are drawn, implied that alteast there are 2 coins of 5 rupee each in the purse.Thus there were 3 cases
a) all 5 rupee coin
b) 3 coins of 5 rupee and 1 coin of 1 rupee
c) 2 coins of 5 rupee and 2 coins of 1 rupee

Now why probability of drawing 2 5rupees coins were obtained?
also how come its 1, 1/2 and 1/6 respectively for case a, b,and c.
should it not be 1/3, 1/4, 1/6 respectively.

Further, I didnt understand how we have calculated the probablilty of case a, b & c.

3) In a question as "If we throw a dice 10 times what is the probability of it showing 6 exactly 3 times?"
Should not the solution be (1/6)^3 * (5/6)^7.
Why was it needed to found "ways of those 3 out of 10" i.e 10c3 and getting its product

 

Re: Probability
by Harsh Aggarwal - Tuesday, 28 August 2012, 08:42 PM
 

Hi ,
Approach :- I feel by keeping 2 at first place and arranging the rest of 9 even numbers in any order will give the required answer.

Favourable case is 2 at first place.So no. of favourable case is 9! for keeping 2 at first place .

And total case is 10 ! for 10 even numbers.  

 Prob  =  9!/ 10 !  = 1/10 .

Hence your answer is correct.

Re: Probability
by DHEERAJ KUMAR - Thursday, 30 August 2012, 06:54 PM
  Hi Rythm
lets see what happens in case a)
the probability that the first coin drawn is 4/4 =1
and the probability that the second coin is also a 5 rupee coin is 3/3=1 ....
therefore , the probability that in two draws it is a 5 rupee coin is 1*1=1.

b)
for this case the probability of getting 5 rupee coin in first draw is 3/4
and for the second draw it is 2/3(as the coin drawn is the last case is not replaced)
therefore the probability of drawing two five rupee coin in this case is (3/4)*(2/3)=1/2

c) similarly for the first draw probability in this case is 2/4 and for the second case it is 1/3
and therefore the for two cases the probability is (2/4)*(1/3)=1/6

I hope this puts to rest all your doubts.... smile smile smile
Re: Probability
by game starts - Saturday, 15 September 2012, 06:43 PM
  kiss
Re: Probability
by anand Ma - Saturday, 11 May 2013, 10:40 PM
  Hi

how we get the probaility that case exist as 1,1/2,1/6
dnt get the ans by dheeraj

Re: Probability
by anand Ma - Monday, 13 May 2013, 12:52 PM
 

Can any anyone explain it again..

As here dheeraj is talking about drawing coin but i think here we have to fing probability of the cases .As we know that there are already  2 5rs coin so there are 3 cases

1)2 (5rs coin) 2 (1rs coin)

2) 3 (5rs coin) 1 (1rs coin)

3) 4(5rs coin)

 

so i guess, we have to pay attention on the 3rd and 4th coin only.

Re: Probability
by Dinesh K - Tuesday, 11 June 2013, 12:02 PM
  it should be 1/3 * 3/5 (prob of choosing a bag * prob of choosing red from bag 1)
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Re: Probability
by amardeep singh - Thursday, 18 July 2013, 06:18 PM
  EXACTLY CORRECT
Re: Probability
by Shashank Dave - Thursday, 15 October 2015, 06:56 PM
  I cannot really understand one thing in the coin question(One in which minimum number of 5 rupees coins is 2), why have we calculated the possibility of existence of each of the cases as different? aren't all the three cases equally likely? Why are they not? Please explain its killing me!
Re: Probability
by Purnendu sharma - Wednesday, 28 October 2015, 08:31 PM
  Hye TG Sir, I wanted to know in d bag,pockets and coin transfer question aftr getting CASE 1:2/5*1/3=2/15 ;CASE2: 3/5*1/2 ;CASE3:3/5*1/2 ;CASE4 :2/5*2/3.
U said sir to sum all these..doing so total prob coming=1..then u said sir your ans is coming 1/2
Is 1/2 the ultimate ans as asked or are there any more steps?? If yes 1/2 ultimate ans,then we dont need to do so lengthy process..1/2 is coming because we are playing on the intuition of either gold or silver coin.so prob of each has to be 1/2..but then what about all these data..
What I think is correct is as follows:
Total prob=(prob of case 1/prob of all cases)*prob of selection of silver coin frm case 1
+ (prob case2/prob of all cases)*prob of selctn silvr coin frm cse 2 + and so on till case 4
Notesad1) (prob of all cases)is coming to be nd has to be 1. (2) prob of selectn of silver coin as multiplied above is d prob of selectn of silver coin aftr d transfer is done
So we get total prob=(2/15*1/3)+(3/10*1/2)+(3/10*1/2)+(4/15*2/3)=47/90
Thus total prob has shifted 4m 1/2 to 47/90 in the given circumstances...
please sir kindly let us know.. I actually found a question with this approach..i understd this apprch nd thpt it wud apply here too..Thanc for the entire material TG sir
Probability
by ashish sharma - Sunday, 7 February 2016, 05:10 PM
  Hey TG!!!!
Plz clarify one of mine doubt. In a question, where 2 cards are drawn without replacement, the probability of a jack and queen has been answered using probability multiplication i.e. P(A) X P(B).

However, I am unable to understand how can we use probability multiplication since both are dependent events and the very basic definition for using this rule states that the events shall be independent of each other.

Plz clarify?
Re: Probability
by ashish sharma - Sunday, 7 February 2016, 06:19 PM
  In addition one more query in the example of dice thrown 10 times and probability of it showing "six" exactly 3 times?

If I go by conventional method, total outcomes= 6^10
Favorable outcomes= 10!/(3!X7!)
However, as solved by you the probability is way higher than above. So, where exactly I am missing additional favorable cases?
Please clarify