M28-27

Question

Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?

A.  \frac{7}{8}

B.  \frac{3}{4}

C.  \frac{2}{3}

D.  \frac{5}{8}

E.  \frac{3}{8}

Bunuel · Accepted Answer

Consider this: there are 8 cards, 5 spades, and 3 hearts. What is the probability that the first card you pick will be a spade? Clearly 5/8. Now, the next question: suppose I throw away three cards at random, not telling you which cards I have thrown away. What is the probability NOW that you pick a spade out of the 5 remaining cards? I've just reduced the sample from which you pick, but does the probability change? WHY should it change? Another question, using the same cards (5 spades and 3 hearts): I say that you can pick one card, but only out of randomly selected 5 cards. What is the probability that the card you pick will be a spade? There is no difference in ALL the above cases, and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking a heart, on the other hand, decreased (or increased). Why would it? Similar questions to practice: https://gmatclub.com/forum/a-box-contai ... 90272.html https://gmatclub.com/forum/a-bag-contai ... 00023.html https://gmatclub.com/forum/each-of-four ... 01553.html https://gmatclub.com/forum/if-40-people ... 97015.html https://gmatclub.com/forum/a-bag-contai ... 00023.html Hope this helps.

Bunuel · Answer

Official Solution:

Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?

A.  \frac{7}{8}

B.  \frac{3}{4}

C.  \frac{2}{3}

D.  \frac{5}{8}

E.  \frac{3}{8}

Basically, we need to find the probability that the seventh marble drawn is red (so, not blue).
 
Now, the initial probability of drawing a red marble is  \frac{5}{8} . Without knowing the other results, the probability of drawing a red marble will not change for ANY successive draw: the second, third, fourth, ..., and seventh. Thus, the probability that the seventh marble is red is  \frac{5}{8} .
 
The same applies to the blue marble: the probability of drawing a blue marble is  \frac{3}{8} . The probability that, for instance, the 8th marble drawn is blue is still  \frac{3}{8} . There is simply no reason to believe WHY any draw would be different from another (provided we don't know the other results).

Answer: D

francisdarby · Answer

Hi there, A new member here, just joined a day or two back

I'm quite confused about the solution to this problem. How is it possible that the probability remains constant even when we are removing marbles without replacement? Wouldn't the denominators in the probabilities adjust for each draw? Any help would be greatly appreciated! Thanks, Francis

spence11 · Answer

For those of you who are confused by Bunuel's beautiful answer, you can arrive at the same solution using permutations.

Let's find the possible permutations for the first 6 balls, and then worry about the 7th and 8th after that

5 Reb and 1 Blue - We are arranging 6 balls of two different types. To arrange 6 objects without respect for repetition of each type, the formula is simply  6! . To account for repeating types, reduce  6!  by the factorial of the repetition number for each type.

\frac{6!}{5!1!}=6

4 Red and 2 Blue

\frac{6!}{4!2!}15

3 Red and 3 Blue

\frac{6!}{3!3!}=20

Now multiply each by the available remaining permutations of the 7th and 8th ball:

5 Reb and 1 Blue?

Only BB is left, so  6*1=6

4 Reb and 2 Blue?

RB and BR are left, so  15*2=30

3 Reb and 3 Blue?

Only RR is left, so  20*1=20

Thus, there are  6+30+20=56  total arrangements

In how many of these last three options are the 7th ball red?

5 Red and 1 Blue:  0 
4 Red and 2 Blue:  15 
3 Red and 3 Blue:  20

\frac{35}{56}=\frac{5}{8}

Answer D

Zksgmat · Answer

A simple answer using combinatorics

if the 7th slot is NOT BLUE, it is RED
now ignore this slot completely. it will not affect the other arrangements. for the 7 remaining slots, we have 7 marbles = (7! arrangements) but since 4 Red repeats and 3 blue repeats, the possible arrangements = 7!/(3!4!) = 35
total arrangements without the restriction for the 7th slot = 8!(5!3!) = 56
desired probability = 35/56 = 5/8

bprashanth88 · Answer

I think this is a  high-quality  question. I have an alternate solution, 8!/(5!*3!) - Total # of permutations - 56 -----(i)

Of these fixing the 7th position as red - # of permutations is 
7!/(4!*3!) - 35------(ii)

(ii)/(i) is what we want = (5/8)

NonPlus · Answer

I used a different method using seating arrangements.
Suppose there are 8 baskets placed in linear fashion, with each basket capable of holding only 1 ball. Suppose we place 8 balls 1-by-one in each basket 
Then, total number of such arrangements - 8!/5!*3! = 56

Now, fix 7th basket with red ball placed in that. Now, we have to find number of arrangements of 7 balls with 4 red balls and 3 blue balls 
==> 7!/4!*3! = 35

Probability = Number of desired outcomes/Total Outcomes = 35/56 or 5/8

SDW2 · Answer

Hello  Bunuel ,  VeritasKarishma  ,  IanStewart 
Even I have solved using this method. Is this approach correct?

KarishmaB · Answer

Yes, it is. 
Or think about it assuming every element is distinct (it doesn't matter in probability). Say every marble is numbered from 1 to 8 (5 are red, 3 are blue). When you pick one after the other without replacement, it's like giving them a rank from 1 to 8. 
So you can arrange 8 items in 8 slots in 8! ways. Every element is equivalent. So each of the 8 elements will appear in the 7th spot in 1/8th of total cases. 
So 5 Red will appear in the 8th spot in 5/8th of total cases.

Hence, probability is 5/8

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Tukkebaaz · Answer

This question is a classic example of a question having multiple correct approaches to solve. I see many folks have highlighted equally useful approaches above adding my 2 cents:

So the Qn says 7th slot as not blue so has to be red. The possible combos then are: 3R3B RR (or) 4R2B RB (we can also have BR at the end in this combo but that is not needed as per Qn so ignored it).

The prob curve is normally distributed, said differently Prob of 1st slot = prob of nth slot, 2nd slot - (n-1)th slot - this logic/ concept can be used widely in other questions as well - Ex: Probability of getting certain dice roll sums, coin tosses results so on, coefficient of expansion of (1+x)^n type questions and so on .

The application here is Prob required is RR + RB   = (5*4/8*7) + (5*3/8*7) =  5/8 (Option D)

Josvit · Answer

nice explanation and reasoning. When solving, I realized that for the second the probability won´t change and neither will for the third and so on. But I wasn´t sure, with this, all clear!

SudsMeister · Answer

How I went about it:
The 7th marble cannot be blue. Thus, the 7th marble has to be red.
The 8th marble can be blue or red.

Case 1: When 7th marble is red, 8th marble is blue. _ _ _ _ _ _ R B
Remaining marbles are (5 - 1) = 4 red and (3 - 1) = 2 blue. The 1st six spaces can be filled by 4 red and 2 blue marbles in  \frac{6!}{4! * 2!}  ways = 15 ways.

Case 2: When 7th marble is red, 8th marble is also red. _ _ _ _ _ _ _ R R
Remaining marbles are (5 - 2) = 3 red and (3 - 0) = 3 blue. The 1st six spaces can be filled by 3 red and 3 blue marbles in \frac{ 6!}{3! * 3!}  ways = 20 ways.

Total possible ways to pick 8 marbles where 5 are red and 3 are blue =  \frac{8!}{5! * 3!}  = 56 ways.

Required probability = (Case 1 + Case 2) / Total ways =  \frac{15 + 20}{56} = \frac{35}{56 }= \frac{5}{8}

Answer D.

Paattaa · Answer

I like the solution - it’s helpful.

Mahi2804 · Answer

I did not quite understand the solution. The questions mention without replacement so how can we assume that the probability remains the same

Bunuel · Answer

“Without replacement” changes conditional probabilities after you see earlier draws. But before you see them, the 7th position is just a random spot in a random permutation, so its chance of being red equals the initial red fraction: 5 out of 8.

Please review the discussion above and check the links provided for similar questions to better understand the topic. Hope this helps clarify your doubt.

NamanMMMM · Answer

I like the solution - it’s helpful. This one seems good to me.

GMATFE2025 · Answer

If you can't think the way Bunnel thinks of this;

5R3B

Total no of ways to arrange RRRRRBBB = 8!/(5!*3!) = 56 ways

7th shouldn't be blue means it is a Red. So fix the 7th position to R and then possible ways = RRRRBB(R)B= 7 positions left that can be arranged in = 7!/(4!*3!) = 35 ways.

Probability = 35/56 = 5/8. (Answer)

Thanks,

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