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The probability that a visitor at the mall buys a pack of [#permalink]
20 Oct 2009, 05:44

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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

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) --> 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.

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 _________________

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 * 2/3 * 1/2 * 2/5 * 1/3

In this case we know that drawing has already happened and the sum of two cards was 8.

What combinations of two cards are possible to total 8? (first card, second card) (6,2) (2,6) (5,3) (3,5) (44) – only 5 possible ways sum to be 8. One from this 5 has already happened.

From this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.

Answer: D (2/5)

If the problem were: 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. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5? (5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36. P(one card 5, sum 8)=2/36=1/18 _________________

0.3*0.3*0.7 is equivalent to saying the the first person picks candy and second person picks candy and third person doesn't pick candy.

however this is a different case compared to 0.3*0.7*0.3 is equivalent to saying the the first person picks candy and second person doesn't pick candy and third person picks candy.

which in turn is a fifferent case compared to 0.7*0.3*0.3 (I'll skip the verbose comments)

Hence the answer is 3*(0.3*0.3*0.7)=0.189

Answer to the second question -

Probability of picking a sibling from the junior class is 60/1000 and Probability of picking the corresponding sibling from the senior class is 1/800. However in this case if we flip the case we still end up with the same sibling pair since the order is not important. Hence we don't multiply by 2 Answer is 3/40,000.

Answer to the third question - We just list the various cases where the sum will be 8

6 2 2 6 5 3 3 5 4 4

Total = 5 5 occurs in 2 cases. hence prob is 2/5. _________________

Problem 1: As there are three visitors, scenarios can be as below YYN,YNY,NYY(Y=buy, N=didn't buy) so Probability= (3/10*3/10*7/10)+(3/10*7/10*3/10)+(7/10*3/10*3/10) = 0.189

Problem 2: Probability of selecting 1 sibling pair= (60/1000*60/800*1/60*1/60)= 1/800000 This can happen 60 times so probability= (1/800000)*60= 3/40000

Problem 3: possible pairs are (2,6),(3,5),(4,4),(5,3),(6,2) so total 5 possibilities two pairs are having 5 so probability = 2/5

This interpretation is wrong. Probability = Favourables / Total. But this question total is constrained if you read it carefully - "the sum of two cards would be 8". Total is not the absolute ways in which you will pick the cards i.e. 6 * 6 = 36 absolute ways. Total ways to get a sum of 8 using two cards is 5. And that's why the ultimate probability is 2/5

Chetangupta wrote:

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. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5? (5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36. P(one card 5, sum 8)=2/36=1/18

Can you please explain why number of outcomes will be 36?

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 * 2/3 * 1/2 * 2/5 * 1/3

In this case we know that drawing has already happened and the sum of two cards was 8.

What combinations of two cards are possible to total 8? (first card, second card) (6,2) (2,6) (5,3) (3,5) (44) – only 5 possible ways sum to be 8. One from this 5 has already happened.

From this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.

Answer: D (2/5)

If the problem were: 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. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5? (5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36. P(one card 5, sum 8)=2/36=1/18

Hi Bunuel,

Thanks for the great explanations. I'm stuck on #3 -- first two worked out great.

-In the scenario where we get 2/5, why isn't it 2/10 because isn't the favorable outcome(2) and the total outcomes is the total number of cards, which is 10? therefore 2/10?

-Can you please talk about the second scenario a little bit. I'm not catching the nuance that makes you multiply 6*6? Why wouldn't it be 6!?

The number of total outcomes is 5, not 10: (6,2) (2,6) (5,3) (3,5) (4, 4). Only 5 possible ways sum to be 8.

So, the probability that one of the cards drawn was a 5 is 2/5.

As for your second question: 6 options for the first card and 6 options for the second card, thus total 6*6=35 options. _________________

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. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5? (5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36. P(one card 5, sum 8)=2/36=1/18

Can you please explain why number of outcomes will be 36?

Regarding the below listed explanation from funnel -

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

So basically you are considering the case when the first sibling selected is from the junior class and the second sibling is from the senior class.....but why does it have to be only in that order.....why would you not consider the case when first sibling is from senior class and the second is from the junior class?

Please explain your logic for considering that case?

1. (0.3)(0.3)(0.7) x 4C3 = 0.189 2. 60/1000 x 1/800 = 3/20000 _________________

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Re: The probability that a visitor at the mall buys a pack of [#permalink]
21 Sep 2013, 11:02

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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 * 2/3 * 1/2 * 2/5 * 1/3

In this case we know that drawing has already happened and the sum of two cards was 8.

What combinations of two cards are possible to total 8? (first card, second card) (6,2) (2,6) (5,3) (3,5) (44) – only 5 possible ways sum to be 8. One from this 5 has already happened.

From this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.

Answer: D (2/5)

If the problem were: 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. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?

What combinations of two cards are possible to total 8 AND one of them to be 5? (5,3) (3,5) – two favorable outcomes.

Total number of outcomes=6*6=36. P(one card 5, sum 8)=2/36=1/18

Hi Bunuel,

Thanks for the great explanations. I'm stuck on #3 -- first two worked out great.

-In the scenario where we get 2/5, why isn't it 2/10 because isn't the favorable outcome(2) and the total outcomes is the total number of cards, which is 10? therefore 2/10?

-Can you please talk about the second scenario a little bit. I'm not catching the nuance that makes you multiply 6*6? Why wouldn't it be 6!?

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