- #1
akhilnair
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- 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/
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/