Probability / high school math: among 20 visibly identical candies, 6 are strawberry

uneatenbiscuit

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Hello!

Our math teacher gave us a surprise test a few days ago. There was one question on it that I haven't been able to answer and it is bothering me very much. My teacher's response was that my thinking around the question was incorrect. I would appreciate it if someone could help me out with it. The question is as follows (I am recalling this question based on memory):

In a bowl with a total of 20 visibly identical candies, 6 are strawberry flavoured. If I pick 10 at random, what is the probability that at least 1 candy will be strawberry flavoured?

My thinking was that even if we pick 10 at a time, we are still picking from the total number 20. So the probability should be 6/20 or 30%. My answer was incorrect and I wasn't told what the correct answer was. Appreciate any help I can get!
 
Hello!

Our math teacher gave us a surprise test a few days ago. There was one question on it that I haven't been able to answer and it is bothering me very much. My teacher's response was that my thinking around the question was incorrect. I would appreciate it if someone could help me out with it. The question is as follows (I am recalling this question based on memory):

In a bowl with a total of 20 visibly identical candies, 6 are strawberry flavoured. If I pick 10 at random, what is the probability that at least 1 candy will be strawberry flavoured?

My thinking was that even if we pick 10 at a time, we are still picking from the total number 20. So the probability should be 6/20 or 30%. My answer was incorrect and I wasn't told what the correct answer was. Appreciate any help I can get!
These "at least" or "at most" questions can be solved in a quick, but perhaps non-intuitive way and a slow, but perhaps intuitive way. Let's do the slow way first because it is just common sense and the basic rules of combinatorics.

Probability of at least 1 = probability of exactly 1 or exactly 2 or exactly 3 or exactly 4 or exactly 5 or exactly 6. Common sense.

Now as you know the probability of a or b = probability of a + probability of b - probability of a and b. But the probability of exactly x and not x is 0.

So probability of exactly 1 or exactly 2 or exactly 3 or exactly 4 or exactly 5 or exactly 6 =

\(\displaystyle \text {P(exactly } 1 \text {)} + \text {P(exactly } 2 \text {)} + \text {P(exactly } 3 \text {)} + \text {P(exactly } 4 \text {)} + \text {P(exactly } 5 \text {)} + \text {P(exactly } 6 \text {).}\)

This is a bit more than common sense, but it requires no more than remembering the rules for computing the joint probability of mutually exclusive events.

There are six strawberry. What is the number of ways that you can pick exactly x of them?

\(\displaystyle \dbinom{6}{x} = \dfrac{6!}{x! * (6 - x)!}.\)

You learned that when you were learning about combinatorics.

How many non-strawberry are there? 20 - 6 = 14. If you are picking 10 and exactly x are strawberry, how many are non-strawberry?

Obviously 10 - x. Again, common sense.

And how many ways are there to pick 10 - x from 14?

\(\displaystyle \dbinom{14}{10 - x} = \dfrac{14!}{(10 - x)! * \{14 - (10 - x)\}} = \dfrac{14!}{(10 - x)! * (4 + x)!}.\)

Again basic combinatorics. And how many ways are there to pick 10 from 20?

\(\displaystyle \dbinom{20}{10} = \dfrac{20!}{10! * (20 - 10)!} = \dfrac{20!}{(10!)^2}.\)

Therefore the probability of getting x strawberry and 10 - x non-strawberry is

\(\displaystyle \dfrac{\dfrac{6!}{x! * (6 - x)!} * \dfrac{14!}{(10 - x)! * (4 + x)!}}{\dfrac{20!}{(10!)^2}}.\)

Now you can work that out for x = 1, x = 2, x = 3, x = 4, x = 5, x = 6.

\(\displaystyle \text {P(exactly } 1 \text {) } \approx 6.50\%.\)

\(\displaystyle \text {P(exactly } 2 \text {) } \approx 24.38\%.\)

\(\displaystyle \text {P(exactly } 3 \text {) } \approx 37.15\%.\)

\(\displaystyle \text {P(exactly } 4 \text {) } \approx 24.38\%.\)

\(\displaystyle \text {P(exactly } 5 \text {) } \approx 6.50\%.\)

\(\displaystyle \text {P(exactly } 6 \text {) } \approx 0.54\%.\)

If you add those up, you get

\(\displaystyle \text {P(at least } 1 \text {) } \approx 99.46\%.\)

That is very straightforward logically, but it is very time consuming.

The other way is to say that the probability of none plus the probability of at least one = 1. So the probability of at least one is 1 minus the probability of none.

\(\displaystyle 1 - \dfrac{\dfrac{6!}{0! * (6 - 0)!} * \dfrac{14!}{(10 - 0)! * (4 + 0)!}}{\dfrac{20!}{(10!)^2}} = 1 - \dfrac{6!}{6!} * \dfrac{14!}{10! * 4!} * \dfrac{10! * 10!}{20!} =\)

\(\displaystyle 1 - \dfrac{14!}{4!} * \dfrac{10!}{20}! =\)

\(\displaystyle 1 - \dfrac{14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5}{20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11} \approx \)

\(\displaystyle 1 - 0.0054 = 100.00\% - 0.54\% = 99.46\%.\)

Two different ways to get to the same result.
 
Last edited:
Hello!

Our math teacher gave us a surprise test a few days ago. There was one question on it that I haven't been able to answer and it is bothering me very much. My teacher's response was that my thinking around the question was incorrect. I would appreciate it if someone could help me out with it. The question is as follows (I am recalling this question based on memory):

In a bowl with a total of 20 visibly identical candies, 6 are strawberry flavoured. If I pick 10 at random, what is the probability that at least 1 candy will be strawberry flavoured?

My thinking was that even if we pick 10 at a time, we are still picking from the total number 20. So the probability should be 6/20 or 30%. My answer was incorrect and I wasn't told what the correct answer was. Appreciate any help I can get!
I'd like to show you why your way is not correct. I agree that 6 out of 20 candies are strawberry and that 6 out of 20 is the same as 6/20 or 30%. Good work here! Your answer to the problem of 30% does NOT take into account that you pick 10 candies. If for example you pick all 20 candies then for certain you would have picked at least 1 strawberry candy so the probability would be 1 or 100%. On the other hand, if you just picked 1 candy, then the probability that you picked at least 1 strawberry candy ( you will get 0 or 1 strawberry candy) would be 6/20. So the probability you are looking for does take into account how many candies you choose!
 
I'd like to show you why your way is not correct. I agree that 6 out of 20 candies are strawberry and that 6 out of 20 is the same as 6/20 or 30%. Good work here! Your answer to the problem of 30% does NOT take into account that you pick 10 candies. If for example you pick all 20 candies then for certain you would have picked at least 1 strawberry candy so the probability would be 1 or 100%. On the other hand, if you just picked 1 candy, then the probability that you picked at least 1 strawberry candy ( you will get 0 or 1 strawberry candy) would be 6/20. So the probability you are looking for does take into account how many candies you choose!

Thank you for your response. I understand the mistake I was making in the thinking/setup of the problem. :D
 
These "at least" or "at most" questions can be solved in a quick, but perhaps non-intuitive way and a slow, but perhaps intuitive way. Let's do the slow way first because it is just common sense and the basic rules of combinatorics.

Probability of at least 1 = probability of exactly 1 or exactly 2 or exactly 3 or exactly 4 or exactly 5 or exactly 6. Common sense.

Now as you know the probability of a or b = probability of a + probability of b - probability of a and b. But the probability of exactly x and not x is 0.

So probability of exactly 1 or exactly 2 or exactly 3 or exactly 4 or exactly 5 or exactly 6 =

\(\displaystyle \text {P(exactly } 1 \text {)} + \text {P(exactly } 2 \text {)} + \text {P(exactly } 3 \text {)} + \text {P(exactly } 4 \text {)} + \text {P(exactly } 5 \text {)} + \text {P(exactly } 6 \text {).}\)

This is a bit more than common sense, but it requires no more than remembering the rules for computing the joint probability of mutually exclusive events.

There are six strawberry. What is the number of ways that you can pick exactly x of them?

\(\displaystyle \dbinom{6}{x} = \dfrac{6!}{x! * (6 - x)!}.\)

You learned that when you were learning about combinatorics.

How many non-strawberry are there? 20 - 6 = 14. If you are picking 10 and exactly x are strawberry, how many are non-strawberry?

Obviously 10 - x. Again, common sense.

And how many ways are there to pick 10 - x from 14?

\(\displaystyle \dbinom{14}{10 - x} = \dfrac{14!}{(10 - x)! * \{14 - (10 - x)\}} = \dfrac{14!}{(10 - x)! * (4 + x)!}.\)

Again basic combinatorics. And how many ways are there to pick 10 from 20?

\(\displaystyle \dbinom{20}{10} = \dfrac{20!}{10! * (20 - 10)!} = \dfrac{20!}{(10!)^2}.\)

Therefore the probability of getting x strawberry and 10 - x non-strawberry is

\(\displaystyle \dfrac{\dfrac{6!}{x! * (6 - x)!} * \dfrac{14!}{(10 - x)! * (4 + x)!}}{\dfrac{20!}{(10!)^2}}.\)

Now you can work that out for x = 1, x = 2, x = 3, x = 4, x = 5, x = 6.

\(\displaystyle \text {P(exactly } 1 \text {) } \approx 6.50\%.\)

\(\displaystyle \text {P(exactly } 2 \text {) } \approx 24.38\%.\)

\(\displaystyle \text {P(exactly } 3 \text {) } \approx 37.15\%.\)

\(\displaystyle \text {P(exactly } 4 \text {) } \approx 24.38\%.\)

\(\displaystyle \text {P(exactly } 5 \text {) } \approx 6.50\%.\)

\(\displaystyle \text {P(exactly } 6 \text {) } \approx 0.54\%.\)

If you add those up, you get

\(\displaystyle \text {P(at least } 1 \text {) } \approx 99.46\%.\)

That is very straightforward logically, but it is very time consuming.

The other way is to say that the probability of none plus the probability of at least one = 1. So the probability of at least one is 1 minus the probability of none.

\(\displaystyle 1 - \dfrac{\dfrac{6!}{0! * (6 - 0)!} * \dfrac{14!}{(10 - 0)! * (4 + 0)!}}{\dfrac{20!}{(10!)^2}} = 1 - \dfrac{6!}{6!} * \dfrac{14!}{10! * 4!} * \dfrac{10! * 10!}{20!} =\)

\(\displaystyle 1 - \dfrac{14!}{4!} * \dfrac{10!}{20}! =\)

\(\displaystyle 1 - \dfrac{14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5}{20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11} \approx \)

\(\displaystyle 1 - 0.0054 = 100.00\% - 0.54\% = 99.46\%.\)

Two different ways to get to the same result.

Thank you for your response.

I haven't yet learnt the mathematical way of solving these types of questions in school as you have demonstrated. I am surprised that our teacher presented us this question. Maybe they were just looking for our approach to such a question.

However, I understand both the approaches you have taken to solve the problem. They seem pretty straightforward and easy to follow.

Appreciate the help! :D
 
I'd like to show you why your way is not correct. I agree that 6 out of 20 candies are strawberry and that 6 out of 20 is the same as 6/20 or 30%. Good work here! Your answer to the problem of 30% does NOT take into account that you pick 10 candies. If for example you pick all 20 candies then for certain you would have picked at least 1 strawberry candy so the probability would be 1 or 100%. On the other hand, if you just picked 1 candy, then the probability that you picked at least 1 strawberry candy ( you will get 0 or 1 strawberry candy) would be 6/20. So the probability you are looking for does take into account how many candies you choose!
Excellent.
 
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