Probability problem

I just reviewed the Probability section and I believe I get the concept. However I was presented with a question from the question of the day I receive and couldn’t get it right. Could you let me know what I’m doing wrong?

Question is: A bag contains 3 gold coins and 7 silver coins. The bag contains nothing else. If 3 coins are selected at random and without replacement from the bag, what is the probability that exactly 2 of the coins are silver?

7/40
7/24
3/10
21/40
2/3

I made three fractions with what I want over total possible: 7/10, 6/9, and 3/8 for the first silver, second silver and first gold. I multiplied across to get 126/720 and selected A.

I apologize if this is a silly question or has been answered somewhere else.

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Hey @john1,

I believe the answer would be 21/40.

The process of (S: 7/10) * (S: 6/9) * (G: 3/8) = 126/720 = 7/40 is correct.

However, as you mentioned, this is for calculating SSG. The question doesn’t mention the order, which means that SGS and GSS are also possibilities. All three possibilities have a 7/40 probability so we can do 3 * (7/40) = 21/40.

That’s where I get lost. I figured that even if it was SGS or GSS the method would just be:

(S: 7/10) * (G: 3/9) * (S: 6/8) which would still equal 126/720 = 7/40 for SGS
or (G: 3/10) * (S: 7/9) * (S: 6/8) which would also still equal 126/720 for GSS

I am not sure where the 7/40 comes from.

Thank you for your quick response!

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For each configuration (i.e. SSG, SGS, GSS), there are 126 winning out of 720 possible outcomes.

Because they don’t overlap, we can add them up. SSG outcomes + SGS outcomes + GSS outcomes = total outcomes.

So we have (SSG: 126/720) + (SGS: 126/720) + (GSS: 126/720) = 3 * (126/720) = 378/720.

We don’t really care about the raw number of outcomes though - what we care about is the probability. So it’s a lot simpler to reduce the fraction as much as we can.

126/720 = 0.175
7/40 = 0.175

378/720 = 0.525
21/40 = 0.525

It can be easier to think about it with smaller and more manageable numbers. What’s the probability that if we flip a coin twice, we get exactly one heads? There are 4 possible outcomes: HH, HT, TH, and TT. Of these, 2 are winners: HT and TH. So we have a 2/4 chance of winning. You could simplify that to 1/2, 50%, or 0.50 - it’s all the same value.

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I feel very dumb, and I hope I’m not wasting too much of your time with all this…

I’m not sure what is wrong, and why I don’t understand this. I get that 126/720 reduces to 7/40, and I understand that multiplying it by three gets us to 21/40. But what was my cue to do that instead of just stopping like I did? Which of the 3 1/2 rules am I messing up here? I keep trying to bring it back to the rules from the book, and I can’t get it to make sense.

Like there was an example from the book where we had the “If two employees are to be selected randomly from among the 250 employees on the company roster, what is the probability that both employees will be male scientists” problem and had to (30/250)·(29/249)= 870/62250 = 1.3975. We didn’t multiply 1.3975 by 2, why not there too?

Even with the coin flip question, I get confused when I’m thinking in terms of the rules. I think to myself "if we are flipping twice that is two fractions, I want heads so my two fractions should each be 1/2, it would be ‘or’ since it is exactly one heads we are looking for, so I’ll add the two fractions. But 1/2 + 1/2 = 1 is obviously wrong.

So I guess I’m wondering what is wrong with the way I am thinking about the rules for probabilities? Can you walk me through how I should be using the rules in each of these problems and how I should identify which rules to use in these problems?

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It’s all good, I just didn’t fully understand your comment!

The trick to this problem is realizing that it is actually three separate and mutually exclusive variants (i.e. SSG, SGS, and GSS), which we refer to in our text as “levels” or “groups”. It’s most similar to the example of picking X men and Y women from an applicant pool at the end of our chapter on Cominatrix rules.

You applied the blank technique correctly for SSG. And if the question was asking what the probability was just for picking in the order of SSG, that would be the answer.

But the question is asking for the probability of two silver and one gold in any order. So, you need to do it again for SGS, and one more time for GSS. And then you can sum up the probabilities for the three variants to get the total probability.

This question does give us a little gift in that the probability for each variant is the same, so we can just multiple by 3 rather than needing to do more complex fraction addition.

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Okay got it, that makes sense. So I needed to realize that there are three variants and combine the rules from the probability section and the cominatrix section.

Last question (hopefully), I understand that this is all because of the order that the coins could be drawn in (variants), but I thought that order didn’t matter for things like marbles, coins, teams, etc. So how can I better identify when the order matters vs when it doesn’t.

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Yep, that’s right. As for identifying whether the order matters or not though… I think it’s mostly just carefully reading the question.

Explicitly writing out the “winning” combinations might be a helpful habit.

A bag contains 3 gold coins and 7 silver coins. The bag contains nothing else. If 3 coins are selected at random and without replacement from the bag, what is the probability that exactly 2 of the coins are silver?

Winners: SSG, SGS, GSS

A bag contains 3 gold coins and 7 silver coins. The bag contains nothing else. If 3 coins are selected at random and without replacement from the bag, what is the probability that all three are silver?

Winners: SSS

A bag contains 3 gold coins and 7 silver coins. The bag contains nothing else. If 3 coins are selected at random and without replacement from the bag, what is the probability that the coins are selected in the order: silver, silver, gold?

Winners: SSG

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This has helped a lot, I appreciate your time Justin!

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