Combinatorics: What to Do with "Forbidden Choices"
There are 10 books on a shelf: 5 English books, 3 Spanish books, and 2 Portuguese books. What is the probability of choosing 2 books in different languages?
Incorrect.
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__Alternative Method__:
Probability is the number of wanted results out of the number of total results possible.
>$$\displaystyle \text{Probability} = \frac{\text{Wanted outcomes}}{\text{Total outcomes}}$$
The number of total options to pick 2 books out of 10 (not ordered) is
>$$\text{Total options} = \frac{10!}{(10-2)!2!} = 45$$.
To easily calculate the number of wanted results, subtract the __unwanted__ results from this total. The number of unwanted results is the number of ways to select 2 books in the same language. This depends on what that language is: 2 English books _or_ 2 Spanish books _or_ 2 Portuguese books.
* Two English (pick 2 out of 5): $$\text{Options} = \frac{5!}{(5-2!)2!} = 10$$.
* Two Spanish: $$3$$ options (it's like picking the unwanted book out of 3 possibilities).
* Two Portuguese: $$1$$ option (there are only 2 books in that language).
So the unwanted options are $$10+3+1 = 14$$, which makes the good options $$45-14 = 31$$.
Therefore, the probability is
>$$\text{Total probability} = \frac{31}{45}$$.
Incorrect.
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Incorrect.
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Incorrect.
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Correct.
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Calculate the probability of each scenario:
1) There are 5 English books out of 10 total books, so the probability of picking an English book is $$\frac{5}{10}$$. There are 5 books not in English, so the probability of pulling a second book not in English is $$\frac{5}{9}$$ (there are only 9 books in total because one is already out after the first pick). So the probability of this scenario is
>>$$\text{Eng / Not-Eng} = \frac{5}{10} \times \frac{5}{9} = \frac{25}{90}$$.
2) The probability that the first is Spanish is $$\frac{3}{10}$$, and the probability that the second is not Spanish is $$\frac{7}{9}$$:
>>$$\text{Sp / Not-Sp} = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90}$$.
3) The probability that the first is Portuguese is $$\frac{2}{10}$$, and the probability that the second is not Portuguese is $$\frac{8}{9}$$:
>>$$\text{Port / Not-Port} = \frac{2}{10} \times \frac{8}{9} = \frac{16}{90}$$.
Since there's an "or" relationship between scenarios, add them to get the total probability:
>$$\text{Total probability} = \frac{25}{90} + \frac{21}{90} + \frac{16}{90} = \frac{62}{90} = \frac{31}{45}$$.
$$\frac{3}{10}$$
$$\frac{14}{45}$$
$$\frac{1}{3}$$
$$\frac{31}{90}$$
$$\frac{31}{45}$$
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