Well is the OA C?

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.

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

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.

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!

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.