While this question looks quite daunting, there is a way to solve it in less than 2 minutes. See if you can do so. And don’t worry, even if you can finish it in less than 4 minutes that’s not too shabby!

A magical leather pouch contains 2 alabaster marbles, 3 cerulean marbles, and 5 magenta marbles. First, four marbles are removed at random, without replacement. Next, three more marbles are removed at random, without replacement. What is the probability that the remaining marbles are the same color?

(A) 1/7

(B) 1/8

(C) 1/120

(D) 7/120

(E) 11/120

Got your answer? Take a look at the explanation to see how you did.

E. 11/120

That’s it!

The Answer is (C) i.e 1/120

The explanation is:

The case for alabaster marbles doesn’t exist since, there are 2 of them and we want to check for conditions where the remaining three marbles are all of the same color.

There are in all 7 marbles drawn without replacement. The probability that the 7 marbles drawn out of 10 in the pouch are not cerulean is 1/120.

This implies that the marbles that remained were all cerulean in color.

The probability that the 7 marbles drawn out of 10 in the pouch are not magenta is 0 since the number or magenta marbles is 5 out of a total of 10. So when 7 are drawn, at least 2 of them will definitely be magenta.

Hence the total probability of getting same color marbles is 1/120

Hi Divya,

How do we know that when seven are drawn, 2 of them will definitely be magenta. Can’t we draw 3 cerulean marbles, 2 alabaster marbles, and 2 magenta marbles?(meaning that the remaining 3 marbles are all magenta).

Have another go. If you get stuck, the correct answer and explanation is provided by a few of the comments below.

Good luck 🙂

The answer is E.

The total number of possible outcomes is 10 C 7 which is 120, which is the denominator. For the numerator. There are 2 possibilities: if no cerulean marble is selected which is 1 way and if all alabaster and all cerulean are picked along with 2 magenta. And 2 magentas can be picked in 5 C 2 ways which is 10. So numerators are 1 or 10, which is 1/120 + 10/120 which equals 11/120 E. 🙂

Perfect!

Nice succinct way of explaining it, too!

The answer is E.

The total number of possible outcomes is 10 C 7 which is 120, which is the denominator. For the numerator. There are 2 possibilities: if no cerulean marble is selected which is 1 way and if all alabaster and all cerulean are picked along with 2 magenta. And 2 magentas can be picked in 5 C 2 ways which is 10. So numerators are 1 or 10, which is 1/120 + 10/120 which equals 11/120 E. 🙂

BTW Chris. You’re amazing. You’re one of those people I hope to meet one day. You’re really making a difference with this initiative. And hope I got that right!

Wow, thanks for the kudos. I’m just happy that I’m able to help so many students. And I’m flattered you hope to me–though I’m probably not that interesting 🙂

(E). Compute 2A, 3C, 5M in all possible sequences, that would be 10!/(2!*3!*5!) = 2520.

If the remaining three are 3C, there will be 2A, 5M having 7!/(2!*5!) = 21 possibilities.

If the remaining three are 3M, there will be 2A, 3C, 2M having 7!/(2!*3!*2!) = 210 possibilities. So the probability is (21+210)/2520 = 11/120.

Interesting, I hadn’t seen it done that way before. But that definitely works!

Focusing on just the marbles in the pouch, meaning using combinations for the marbles inside the pouch, instead of outside it, seems a little faster. Either way, you get the answer in a relatively short amount of time 🙂

Answer is E ?

I think it could be E (11/120).

P (remaining marbles of the same colour) = P(All C) + P(All M).

Whether its taken during the first or second time doesnt matter.

so P(all C) = 1/120

and P(all M) = 10/120

Good, concise solution 🙂

let me just break the ice this time.Can we really make this question as what is probability of drawing 3 marbles of the same colors.then answer would be (E)

That would be 1/12+1/120=11/120.

We need to have all the alabaster marbles.Next, we can have either 3 celurean and 2 magneta marbles or 5 magneta marbles so that 3 magneta or 3 celurean will remain.

This can be done in 2C2 x 3C3 x 5C2 + 2C2 x 5C5 ways (i.e.,) 11 ways.The total probability is 11/ 10C7 . so (E) 11/120

Interesting, I hadn’t seen in done exactly like that before. But that definitely works, both logically and in terms of the correct answer. Good job 🙂

Is the answer E.

3c3/10c3+5c3/10c3=11/20

Ans : D: 7/120

Is it E?

Probability = Possibility of picking 3 with same color/Total Number of possibilities

Possibility of picking 3 with same color = 1(all 3 cerulean) + 5c3(3 magenta) = 11

Total Number of possibilities = 10C3 = 120

Hence 11/120

Yep, that’s the most elegant way of doing it as well. That is, the least math involved. Good job 🙂

7/120

M => Magenta ; C => cerulean

The last 3 marbles could be 3 Magenta or 3 cerulean.

Last 3 M’s can be arranged in 7!/(2!*3!*2!) ways,that is, 210 ways.

Last 3 c’s can be arranged in 7!/(2!*5!) ways, that is, 21 ways.

Total = 231

Total no.of ways all the marbles can be arranged = 10!/(2!*5!*3!) = 2520

Hence P(last 3 picks same) = 231/2520 or 11/120

When doing the above calculations,I realised that P(first 3 picks same) = P(last 3 picks same). With this insight, the problem can be solved in under a min.

P(first 3M’s) + P(first 3 C’s) = (5/10 * 4/9 * 3/8) + (3/10 * 2/9 * 1/8) = 11/120.

I like the last way you approached that. Interesting! Whether it’s the first three or last three marbles doesn’t matter. Cool insight 🙂

Yeah, the positioning of the 3 similar marbles in the 10 picks is moot.

Solution begins backwards :

The only two possibilities where same colored marbles remain –

CCC or MMM, corressponding to these two there are only 2 and 6 ways respectively in which marbles could have been picked.

CCC is achieved by (AMMM + AMM) and (AAMM + MMM) = 2 posibilities

MMM is achieved by similar 6 possibilities

Total 8 No. of events will leave us with desired same color marbles, out of the total no. of ways in which 4 then 3 marbles can be picked from 10.

Answer is 11/120

Solution :For the above result there are two scenarios.

we could include two alabaster marbles and 5 magenta marbles in the 7 marbles that are taken leaving behind 3 cerulean marbles

OR

we could take 2 alabaster marbles ,3 cerulean marbles and two magenta marbles leaving behind 3 magenta marbles.

Number of cases in which the above 2 are possible : 7!/(2!*5!) + 7!/(2!*3!*2!)=231

Total number of cases :10!/(2!*5!*3!)=2520

Probability= 231/2520 =11/120

That’s a great way using the MISSISSIPPI rule. But an even more succinct way is using the combination formula to find how many ways you can select for magenta or cerulean (3C3 and 5C3, respectively) out of a 10C3.

Either way, both solutions work!

This problem is same as selected as 3 marbles and all three are of same color.

Since there are only 2 alabaster marbles, we cannot select 3 alabaster marbles, so probability is 0. even mathematically we can calculate it as (2/10)*(1/9)*(0/8) which is equal to 0.

Now for the 3 cerulean marbles, we get probability as (3/10)*(2/9)*(1/8) = 6/720 and for the 5 magenta marbles we get it (5/10)*(4/9)*(3/8) = 60/720.

Now adding the three probabilities we get, 0 + 6/720 + 60/720 = 66/720 = 11/120

So answer is E.

Cool, I like that solution based on the insight that it doesn’t matter whether the marbles of the same color are the first out of the pouch or the three remaining in the pouch. Good job 🙂