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.
_________________

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


What are GMAT Club Tests?
25 extra-hard Quant Tests

Find out what's new at GMAT Club - latest features and updates

Third problem:

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
_________________

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


What are GMAT Club Tests?
25 extra-hard Quant Tests

Find out what's new at GMAT Club - latest features and updates

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

Great Explanations and examples - couldn't have been more clear!!

Kudos given for all of them..

Thanks!!

good questions and excellent explanations
_________________

Success is my Destiny

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

I see no argument of multiplying this by two.

Answer: A.

-------------------------------------------------------------------------------------------------------------

I have a question regarding this one -

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?