Events & Promotions
|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
20 Oct 2009, 06:44
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
77% (01:10) correct
23%
(01:08)
wrong
based on 530
sessions
History
Date
Time
Result
Not Attempted Yet
Three for comparison: How do we decide what formula to use? Can someone explain the logic please..
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027
How do we decide whether the solution is
3/10* 3/10 * 7/10 = .063
OR
3/10* 3/10 * 7/10 *3 = 0.189
Likewise the following question:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior.If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15
How do we decide whether the solution is
60/1000 * 1/800 = 3/40,000
OR
60/1000 * 1/800 * 2 = 3/20,000
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?
* 1
* \(\frac{2}{3}\)
* \(\frac{1}{2}\)
* \(\frac{2}{5}\)
* \(\frac{1}{3}\)
This has yet another method to calculate!!
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027
How do we decide whether the solution is
3/10* 3/10 * 7/10 = .063
OR
3/10* 3/10 * 7/10 *3 = 0.189
Likewise the following question:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior.If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15
How do we decide whether the solution is
60/1000 * 1/800 = 3/40,000
OR
60/1000 * 1/800 * 2 = 3/20,000
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?
* 1
* \(\frac{2}{3}\)
* \(\frac{1}{2}\)
* \(\frac{2}{5}\)
* \(\frac{1}{3}\)
This has yet another method to calculate!!
20 Oct 2009, 21:06
Kudos
Bookmarks
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027
Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189
Answer: C.
The point here is not about to multiply or not by 3, the point is to find # of ways favorable scenario to occur: in our case we are asked to find the probability of 2 out of 3 visitors to buy the candy.
3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t.
NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t.
Let’s consider some similar examples:
1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?
The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did.
So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441
NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did.
2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?
At least ONE buys, means that buys exactly one OR exactly two OR exactly three:
P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657
P(B=1) --> 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s)
P(B=2) --> 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s)
P(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.
BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below:
P(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.
3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?
P(B=2)=5!/2!3!*0.3^2*0.7^3
We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.
Hope now it's clear.
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027
Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189
Answer: C.
The point here is not about to multiply or not by 3, the point is to find # of ways favorable scenario to occur: in our case we are asked to find the probability of 2 out of 3 visitors to buy the candy.
3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t.
NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t.
Let’s consider some similar examples:
1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?
The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did.
So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441
NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did.
2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?
At least ONE buys, means that buys exactly one OR exactly two OR exactly three:
P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657
P(B=1) --> 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s)
P(B=2) --> 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s)
P(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.
BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below:
P(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.
3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?
P(B=2)=5!/2!3!*0.3^2*0.7^3
We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.
Hope now it's clear.
20 Oct 2009, 21:09
Kudos
Bookmarks
Let’s move to the second problem:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15
This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.
What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000
I see no argument of multiplying this by two.
Answer: A.
This problem can be solved in another way and maybe this way shows that no need of multiplication:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Let’s count favorable outcomes: 1 from 60=60C1=60
The pair of the one chosen=1C1=1
So total favorable outcomes=60C1*1C1=60
Probability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000
Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
The same way here:
What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000
Another way:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Favorable outcomes:
1 from 120=120C1=120
The pair of the one chosen=1C2=2
So total favorable outcomes=120C1*1C2=240
Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15
This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.
What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000
I see no argument of multiplying this by two.
Answer: A.
This problem can be solved in another way and maybe this way shows that no need of multiplication:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Let’s count favorable outcomes: 1 from 60=60C1=60
The pair of the one chosen=1C1=1
So total favorable outcomes=60C1*1C1=60
Probability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000
Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
The same way here:
What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000
Another way:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Favorable outcomes:
1 from 120=120C1=120
The pair of the one chosen=1C2=2
So total favorable outcomes=120C1*1C2=240
Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000
