Probability Word Problem: Aptitude Question

[akhilnair]

Homework Statement

Probability Problem: Drawing At Least One Blue Marble in Two Draws Without Replacement

Relevant Equations

In post

Hello everyone,

I’ve been trying to solve this probability problem, but got stuck. Here’s the problem and my attempt:

Problem
In the Squid Game marble challenge, a bag contains 10 red marbles, 8 blue marbles, and 7 green marbles. A player is allowed to draw two marbles without replacement. If the player needs to draw at least one blue marble to proceed to the next round, what is the probability of this happening?

Question: What is the probability that the player draws at least one blue marble in two draws without replacement?
​

My Attempt So Far:
I started by calculating the total number of marbles in the bag:
10 (red)+8 (blue)+7 (green)=25 marbles

The total number of possible outcomes for two draws without replacement is: (25 x 24)/2 = 300

Next, I considered the complement of the event (drawing no blue marbles) to find the probability of drawing at least one blue marble.

1. For the first draw, there are 10+7=17 non-blue marbles out of 25, so the probability of not drawing a blue marble on the first draw is:
P(not blue, first draw)=17/25
2. For the second draw (given no blue marble was drawn first), there are now 16 non-blue marbles remaining out of 24 total marbles:
P(not blue, second draw)=16/24
The probability of drawing no blue marbles in two draws is:
P(no blue)=(17/25) x (16/24) = 272/600

So, the probability of drawing at least one blue marble should be:
P(at least one blue)=1−P(no blue)=1−(272/600)=328/600

But here’s where I’m stuck: I’m not entirely sure if I’ve correctly set up the complement event or whether I’ve made an error in my calculations for the probabilities. Could someone help verify if this approach is correct?

Alternatively, is there a better way to calculate the probability of drawing at least one blue marble directly?

Thank you in advance for your help!

Question source: https://www.learntheta.com/math-problem-squid-game-marble-probability/

 

[Orodruin]

You are correct, but you never used the number of possible outcomes you computed. That belongs to a different way of determining the no blue probability:

You correctly identified (25 C 2) = 300 as the number of ways of drawing 2 of 25 objects.

The number of ways of drawing no blue is the number of ways of selecting 2 of the 17 non-blue, ie, (17 C 2) = 17x16/2 = 136. The remaining 300-136 = 164 combinations have at least one blue so P(at least one blue) = 164/300 = 328/600. Same result.

 

[Gavran]

One more approach:
P(blue on the first draw)=8/25
P(not blue on the first draw, blue on the second draw)=(17/25)x(8/24)
P(at least one blue)=P(blue on the first draw)+P(not blue on the first draw, blue on the second draw)=41/75

 

[PeroK]

But here’s where I’m stuck: I’m not entirely sure if I’ve correctly set up the complement event or whether I’ve made an error in my calculations for the probabilities. Could someone help verify if this approach is correct?

Alternatively, is there a better way to calculate the probability of drawing at least one blue marble directly?

If you can write a computer program, then you can simulate this and check the answer. For elementary problems, it can be a good way to become confident in your methods. In more advanced problems, it is important to check your answers. And, eventually, a computer simulation is the best approach.

You can also simulate things manually by pretending to be a program and seeing what happens if we repeat this experiment many times. In this case, you can imagine repeating the experiment 600 times:

First marble:
Blue 600 x 8/25 = 192 times;
Not blue = 600 x 17/25 = 408 times

You can pause at this point and ask yourself "is that what would really happen". And, hopefully, you would see that yes, that is what would happen. The key point, of course, is that all these 600 events are equally likely.

Then, what if the first marble is blue. We have 7 remaining blue marbles out of 24, so out of the 192 experiments we would expect:

Blue + Blue = 192 x 7/24 = 56
Blue + Not blue = 192 x 17/24 = 136

And, finally, for when the first marble is not blue:

Not Blue + Blue = 408 x 8/24 = 136
Not Blue + Not Blue = 408 x 16/24 = 272

And now you can answer any question. (Although, you would have more cases if you cared about the difference between the red and green marbles).

In this example, we have three cases where at least one blue was drawn, with relative frequency: 56 + 136 + 136 = 328. And, only one case where no blue was drawn: 272.

 

[Steve4Physics]

8Bs and 17NBs. With two draws, there are three possible ways to draw (with no replacement) at least one blue:
(B, NB) or (NB, B) or (B, B).
___________

P(B followed by NB) = $\frac 8{25} \times \frac {17}{24} = \frac {136}{600}$

P(NB followed by B) = $\frac {17}{25} \times \frac 8{24} = \frac {136}{600}$

P(B followed by B) =$\frac 8{25} \times \frac 7{24} = \frac {56}{600}$

P(At least one B is drawn) = $\frac {136+136+56}{600} =\frac {328}{600}$

 

[Orodruin]

If you can write a computer program, then you can simulate this and check the answer.

Anecdotal level evidence shows that some students that think they can write a computer program cannot write a computer program.

Edit: Given the small number of outcomes, simply going through all 600 (or 300 depending …) outcomes is probably both faster than Monte Carlo simulation and without statistical fluctuation.

 

[Orodruin]

I mean, if we are in the habit of doing it every way possible …

Unordered ways of picking two blue (8 C 2) = 28.
Unordered ways of picking one blue and one non-blue (8 C 1) (17 C 1) = 8 x 17 = 136.
Total unordered ways of picking 2 balls (25 C 2) = 300.

P(#blue >= 1) = P(#blue = 2) + P(#blue = 1) = (28+136)/300 = 164/300

Isn’t it amazing how the result comes out the same regardless of how we do it?!

 

[gmax137]

Isn’t it amazing how the result comes out the same regardless of how we do it?!

What are the odds??

 

[Orodruin]

What are the odds??

0, 1/2, or 1 depending on method of computation used. /s