Problem Solving

A bag contains 6 red marbles and 4 blue marbles. If a marble is drawn with replacement, what is the minimum number of draws required so that there is at least a 70% probability of drawing a blue marble.
A. 6
B. 4
C. 3
D. 2
E. 5

To find the probability of drawing a blue marble, we can simply compute 1 - (Probability of drawing all red marbles).

If we only draw one marble, this gives us 1 - (6/10), or 3/5. Not high enough.

If we draw two marbles, this gives us 1 - (6/10)*(5/9), or 2/3. Still not high enough!

If we draw three marbles, this gives us 1 - (6/10)*(5/9)*(4/8), or 5/6. Success!

Hi Matt,
Didn't get how r u getting the probability as 1-(getting all red marble)??Can u pl explain once more?

sandipgumtya wrote:Hi Matt,
Didn't get how r u getting the probability as 1-(getting all red marble)??Can u pl explain once more?

Sure!

When I'm picking these marbles, there are (on some level) only two possible results.

Result #1:: I get all red marbles
Result #2:: I DON'T get all red marbles

Since these are the only two possibilities, we must have

(Result #1) + (Result #2) = 100%

This implies that

Result #2 = 100% - (Result #1), or

Result #2 = 1 - Result #1

From here, we see that

(Getting at least one blue marble) = 1 - (Getting all red marbles)

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


A bag contains 6 red marbles and 4 blue marbles. If a marble is drawn with replacement, what is the minimum number of draws required so that there is at least a 70% probability of drawing a blue marble.
A. 6
B. 4
C. 3
D. 2
E. 5

The probability that at least one marble is blue when drawn out one at a time = 1 - probability that all the marbles are red.

probability that all the marbles are red.

once : 6C1/10C1=4/6=66.7% -->probability that at least one marble is blue when drawn out one at a time= 1-66.7%=33.3%<70%
twice:(6C1/10C1)(6C1/10C1)=9/25=36% -->probability that at least one marble is blue when drawn out one at a time= 1-36%=64%<70%
third times:(6C1/10C1)(6C1/10C1)(6C1/10C1)=27/125=21.6% --> probability that at least one marble is blue when drawn out one at a time= 1-21.6%=78.4%>70%. Thus the answer is 3, C.


If you know our own innovative logics to find the answer, you don't need to actually solve the problem.
www.mathrevolution.com
- The one-and-only World's First Variable Approach for DS and IVY Approach for PS that allow anyone to easily solve GMAT math questions.

- The easy-to-use solutions. Math skills are totally irrelevant. Forget conventional ways of solving math questions.

- The most effective time management for GMAT math to date allowing you to solve 37 questions with 10 minutes to spare

- Hitting a score of 45 is very easy and points and 49-51 is also doable.

- Unlimited Access to over 120 free video lessons at https://www.mathrevolution.com/gmat/lesson

- Our advertising video at https://www.youtube.com/watch?v=R_Fki3_2vO8

thanks Matt.
@Brent:The question says "with replacement"but i think u considered without.can u plz check and confirm.

There's a very easy way to do it.

Consider the following: since the question asks us about the least number of withdraws and replacements we can make in order to achieve 7 marbles we will assume that every time we withdraw a marble it will be red,

now we have the following ratio of red to blue marbles

Red Blue Total
Before making any withdraws 6 4 10
After withdrawing one red marble and switching it with blue 5 5 10
After withdrawing 2 red marbles and changing them to blue 4 6 10
After withdrawing 3 red marbles and changing them to blue 3 7 10

BINGO!

sandipgumtya wrote:Hi Matt,
Didn't get how r u getting the probability as 1-(getting all red marble)??Can u pl explain once more?