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probability questions
by yash chavda - Tuesday, 23 November 2010, 11:50 AM
  (1)If 16 oranges are distributed among 4 children such that each gets at least 3 oranges, the number of ways of distributing them is
a. 30 b. 210 c. 15 d. 35

(2)If all the first 96 multiples of 4 are arranged one after the other the penultimate(second to last in a series or sequence) digit in this sequence is
a. 2 b. 4 c. 8 d. 9
Re: probability questions
by TG Team - Tuesday, 23 November 2010, 02:27 PM
  Hi Yashsmile

1. It is just non-negative number of ways of distribution of 4 identical balls to 4 distinct boxes which is = 7C3 = 35.

2. You just need to find the ten's digit of 96th multiple of 4 which is = 8.

Kamal Lohia
Re: probability questions
by ladybird ldd - Thursday, 27 January 2011, 07:08 PM
  Hello Sir,

I have a doubt here. In the first question, i could infer the following:

16 oranges to 4 people, so that each has AT LEAST 3 oranges.
Let the people be A, B, C , D. Firstly, let's distribute 3 oranges each to all four.
therefore, 4*3 = 12 oranges distributed. Oranges left = 16 - 12 = 4.

Now 4 oranges to be distributed amongst 4 people. hence the number of ways = 4! = 24.

Pls clarify.
Re: probability questions
by TG Team - Friday, 28 January 2011, 05:25 PM
 

My dear ladybird ldd smile

Let's analyse your inference:

4 oranges are to be distributed to 4 people - If I look from oranges' perspective then 1st orange have 4 choices (as it can be distributed to anyone of the four people), now what about 2nd orange? Does it have lesser choices - certainly not. It also has 4 choices and that's also true for remaining two oranges.

Now appears one more doubt that should not answer be 44 then? Look, in our counting we have assumed that all the oranges are different (that's why we are able to name them 1st, 2nd, .. and so on) but actually it's not. So in above counting we have a lot of repetitions which need to be eradicated. And that's a tough task.

So think alternately. We are having 4 identical oranges which are to be distributed to four (certainly) different people. That is, we are to just divide the 4 oranges in four parts and distribute them to four people (say 1st part belongs to 1st person, 2nd part belongs to 2nd person,.. and so on) Now to divide four oranges in four parts, we are to just place 3 partitions anywhere between those 4 oranges. Or we need to select 3 (distinct) places where the partitions will be placed out of total 7 places available (4 for oranges and 3 for partitions).

And this can be easily achieved in C(7, 4) = 7C= 35 ways.

Hope it is clear. smile

Kamal Lohia 

Re: probability questions
by ladybird ldd - Friday, 28 January 2011, 06:17 PM
  Yes sir smile thanks a ton !

Quant is not as draconian as it seems mixed

Re: probability questions
by inesha srivastava - Monday, 5 September 2011, 07:46 PM
  hi Kamal,

Please clarify how you get 7 places to partition 4 oranges into three.
If I look at 4 oranges sitting in which we have to choose three places to partition them, it looks something like

_O_O_O_O_ - so there are five places to choose from to divide 4 oranges with three partitions.



thanks for helping ,
Inesha

Re: probability questions
by TG Team - Wednesday, 7 September 2011, 12:34 PM
 

Hi Inesha smile

Look in this way. I have four oranges OOOO and three partitions ||| i.e. seven objects are to be arranged which will take exactly seven places for their placement.

Now number of ways of arrangement of 4 idential oranges and 3 identical partitions (i.e. OOOO|||) is = 7!/(4!)(3!) = C(7, 3)

This is similar to finding the number of ways to select three places out of these seven where three identical partitions will be arranged in one way and at the remaining four places four identical oranges will be arranged in one way. So these ways of selection are = C(7, 3).

Hope this is clear now. smile

Kamal Lohia

Re: probability questions
by krishna krishna - Monday, 14 November 2011, 08:36 PM
  1.35 ways
2.4
Re: probability questions
by Agnivo Chatterjee - Thursday, 17 November 2011, 06:22 PM
  I think this method can be applies to any similar problem for distributing stuff, thanks sir for clarifying it.

So to make the understanding precise ..

1. Once the number of X fixed to be distributed among "y" take  take the number of ways as 1.

2.The number of ways the the remaining amount to be distributed among y=number remaining+(y-1) c (y-1)

I will be giving it a shot lets see....
Re: probability questions
by Anuj Asthana - Wednesday, 27 June 2012, 05:30 AM
  Kamal Sir I am a Big fan of yours Please help me out with the question -

If we have an empty room, and every week there is a 0.6 probability for a person to join, 0.2 probability for a person to leave, and 0.2 probability for no change (only one action can be performed per week), what is the probability that there are at least 40 people in the room after 104 weeks?

(A) 0.42
(B) 0.55
(C) 0.61
(D) 0.67
(E) 0.74

Please explain how you do it, thanks!

Sir I know that IT would be done by unions and intersections but please guide me and it is some what with atleast 40 in 104 weeks so ?

Sir I also want to join in the CBT club I am dying to join it
Thanks
Regards
Anuj