Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
PS : Probability # 6... History...and a Doubt [#permalink]
16 Sep 2003, 22:22
00:00
A
B
C
D
E
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Hey all
Part 1: We have 4 History, 2 physics and 6 chemisty books. If we draw 6 books, what is the probability that we will have atleast 2 history and 1 chemistry book amongst them?
Part 2 :
Please read the stuff below.
Just want to make sure about this probability concept.. i am a bit confused about this..
There are cases where we multiply the Number of combinations to the probability of events...
Say for example. prob of getting 3 heads in 6 tosses of a die?
1/2 ^3 * 1/2^3 * 6C3
But in some cases ,like the selecting two books of the same color or whatever...we dont take combinations.
Example ...we have 5 History books , 7 Math books..
What is the probability that two books selected are on the same subject?
So we say ..its 5/12*4/11 + 7/12*6/11
Why dont we multiply this by 12C2?
My thoughts: If we have a fixed number of trials and we seek a definite number of favorable outcomes out of those trials, we need to consider all possibilities of the outcomes.
so...3 heads out of 6 tosses can be obtained in 6C3 ways and we know that the prob of a head is 1/2
so take into account all the possbilites, we multiply the 6C3 to the individual event probabilities..
For 5 History, 5 Math Books ..
Since we just select two ...we might as well multiply by 2C2..but that would give us 1. so we effectively dont use the combinations thing in this case..
Do you think i am right...kindly add to this if you please
i can see what u are saying.
praet i always solve probability and combinations use basic fundamentals.
Just to revise:
probability of event X = Number of favourable events that results in X/ Total number of possible outcomes.
Now, two find above two quantities whether u use combinations or permutations really does't matter as far as you know what u are calculating. (i will take your example)
1) we have 5 History books , 7 Math books..
What is the probability that two books selected are on the same subject?
Total outcomes = 12C2 ------------ a
Favourable outcomes = 5C2 + 7C2 --------------- b
2) prob of getting 3 heads in 6 tosses of a die? (errata: read die as coin)
Total outcomes = 2^6 ---------- a
Favourable outcomes = 6C3*1 = 6P3*3! -------- b
Thus, probabiliry = eq(b)/eq(a)
= 6C3/2^6
The point that i am trying to make is that stick to the basic formaula for calculating and follow the counting methods or permutations (which ever applicable to the question asked) to get a and b (as stated above)
and u will be able to solve any probability problem with ease. Its very important not force the solutions of probability questions since a slight variation in actual test may lead to problem/mistakes.
As for your original question again use the above basics:
Part 1
Q: We have 4 History, 2 physics and 6 chemisty books. If we draw 6 books, what is the probability that we will have atleast 2 history and 1 chemistry book amongst them?
Total ways = 12C6 ----------a
Number of favourable ways = 4C2*6C1*9C3 ----------b
Thus probability for above = eq(b)/eq(a)
(Note: i have assumed that all books are different, if books of one type is identical then situation would be different)
hope this clarifies...
- Vicks
errata: For question2: Favourable outcomes = 6C3*1 = 6P3/3! -------- b
i wrote it in my previous mail as: 6P3*3! (Basically when all three items are identical - in this case coin - then we will divide by 3! to get the unique arrangements)
- Vicks
sorry, i should have been more clear...i don't understand the "b" in your explanation for the problem i've pasted at the bottom of this post, i'm clear as to how you reasoned the total possible outcomes, but i don't understand the logic behind >>Favourable outcomes = 5C2 + 7C2 --------------- b
please explain...thanks
>>>Now, two find above two quantities whether u use combinations or permutations really does't matter as far as you know what u are calculating. (i will take your example)
1) we have 5 History books , 7 Math books..
What is the probability that two books selected are on the same subject?
Total outcomes = 12C2 ------------ a
Favourable outcomes = 5C2 + 7C2 --------------- b
The question asks for number of ways of selecting two books on same subject. So out of 5 History books & 7 Math books, i can select two of history in 5C2 ways and two books on Math in 7C2 ways.
Since these events cannot happen together.
Thus total favourable ways = 5C2 + 7C2
point to note: Here i am not mutiplying these ways by 2! to include the order. If i do mutiply then i would have to multiply total ways (12C2) also by 2! but still answer will remain the same..!!
hope this clarifies
-vicky