BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

probablity

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 233
Joined: Mon Jun 09, 2008 1:30 am
Thanked: 5 times

probablity

by blaster » Tue Apr 19, 2011 11:30 pm
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 will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027

[spoiler]I'm calculating it like this 0.3*0.3*0.7 = 0.063 [/spoiler]
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 965
Joined: Thu Jan 28, 2010 12:52 am
Thanked: 156 times
Followed by:34 members
GMAT Score:720

by vineeshp » Tue Apr 19, 2011 11:36 pm
Well is the OA C?
Vineesh,
Just telling you what I know and think. I am not the expert. :)
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 144
Joined: Sun Aug 29, 2010 9:17 am
Thanked: 40 times
Followed by:4 members
GMAT Score:760

by jaymw » Tue Apr 19, 2011 11:49 pm
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 will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027
blaster, your approach is only the first part of the solution. What you calculated would be true of a situation in which the first visitor buys, the second visitor buys and the third visitor does not buy. However, these events don't have to occur in the above order. The question just asks us to calculate the probability that two out of three buy.

To do so, use the formula for binomial distributions:

(n choose k)*p(success)^no. of successes*p(failure)^no. of failures

with:

n=number of visitors in total=3
k=number of successes (visitors who buy)=2
p(success)=p(buy)=0.3
p(failure)=p(no buy)=1-p(buy)=1-0.3=0.7

Now you have:

p(2out of 3 buy)=(3 choose 2)*0.3^2*0.7^1=3*0.09*0.7=0.27*7=0.189

You can always use the above formula when

1. the propabilites of success and failure stay the same
and
2. there are only two outcomes (in this case "buy" and "not buy")

The answer is C.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 436
Joined: Tue Feb 08, 2011 3:07 am
Thanked: 72 times
Followed by:6 members

by manpsingh87 » Wed Apr 20, 2011 1:06 am
blaster wrote: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 will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027

[spoiler]I'm calculating it like this 0.3*0.3*0.7 = 0.063 [/spoiler]
out of 3 customers,any 2 customers that will buy a 2 packets of candy can be selected in 3C2 ways=3,
and the probability of buying the candy pack by selected persons =0.3, also their probability of not buying the candy pack=0.7,
hence required probability=3C2*0.3*0.3*0.7=3*0.3*0.3*0.7=0.189 hence C
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
Top
User avatar
Legendary Member
Posts: 965
Joined: Thu Jan 28, 2010 12:52 am
Thanked: 156 times
Followed by:34 members
GMAT Score:720

by vineeshp » Wed Apr 20, 2011 1:31 am
jaymw wrote:
blaster, your approach is only the first part of the solution. What you calculated would be true of a situation in which the first visitor buys, the second visitor buys and the third visitor does not buy. However, these events don't have to occur in the above order. The question just asks us to calculate the probability that two out of three buy.

To do so, use the formula for binomial distributions:

(n choose k)*p(success)^no. of successes*p(failure)^no. of failures

with:

n=number of visitors in total=3
k=number of successes (visitors who buy)=2
p(success)=p(buy)=0.3
p(failure)=p(no buy)=1-p(buy)=1-0.3=0.7

Now you have:

p(2out of 3 buy)=(3 choose 2)*0.3^2*0.7^1=3*0.09*0.7=0.27*7=0.189

You can always use the above formula when

1. the propabilites of success and failure stay the same
and
2. there are only two outcomes (in this case "buy" and "not buy")

The answer is C.
Hey staying on these lines, do you think it is advantageous to go into binomial distribution n all that for GMAT level probability? If yes, do u know of any good material available online to nail probability. None of the test prep materials go into that detail on probability. Asking you cos u r a 760-er and would know better.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 144
Joined: Sun Aug 29, 2010 9:17 am
Thanked: 40 times
Followed by:4 members
GMAT Score:760

by jaymw » Wed Apr 20, 2011 2:27 am
Well, vineeshp, i might be a bit biased when answering your question because I have a background in teaching/tutoring statistics. That means that I did virtually nothing for GMAT probability questions.

Of course, no GMAT question will ever ask you what kind of a probability distribution underlies the problem. In general, it will be helpful to distinguish between events in which the probabilites change (wihtout repetition) and events in which the probabilities stay the same (with repetition).

I believe that for the GMAT it suffices to know your way around the following 4 concepts:

- tree diagrams
- crosstabs
- binomial distribution
- hypergeometric distribution

The terms seem to be difficult but you don't need to actually know them. Check out wikipedia or some other sources containing info on when to use which concept and you will do fine on GMAT probability!
Top
User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pm
Thanked: 59 times
Followed by:33 members

by fskilnik@GMATH » Wed Apr 20, 2011 5:44 am
blaster wrote: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 will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027
I believe manpsingh87´s approach is the "GMAT-oriented" one!

His solution takes (correctly) into account that:

01. Events A (customer 1 buys candy), B (customer 2 buys candy) and C (customer 3 does not buy candy) are INDEPENDENT, that is, the occurrence of any one of them does not affect the probability of occurring any one of the other two ;

This item validades that Prob(A & B & C) = Prob(A)*Prob(B)*Prob(C) ;

02. All events of the type "exactly 2 of 3 customers buy candy" are EQUIPROBABLE, that is, you could change the customer who does NOT buy candy to be #1 or #2, leaving the other two as buyers.

This item validades that you may get the Prob(A & B & C) value and multiply it by C(3,2) = 3. (This is also related to mutually-exclusive events, feel free to ask me here about it if you wish.)

I hope my observations are useful.

Regards,
Fabio.

P.S.: the correct answer is [spoiler]really (C)[/spoiler], by the way.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Top
Post new topic Post Reply
• Page 1 of 1