# Do You Remember Love?

. — Me@2021-10-30 07:24:36 AM

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2021.10.30 Saturday ACHK

# 機會率驗算 2

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There are 5 women and 6 men in a bus. But there are only 4 seats. Assume that all the seats will be occupied and all possible seating arrangements have the same probability to appear.

What is the probability that 2 or more men will have seats?

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Probability method gets the final probability by multiplying several fractions of probability together.

P method:

This method considers each seat one by one: what is the probability that this seat will get, for example, an M?

Let $\displaystyle{x}$ be the number of men that will have seats. Also, instead of writing $\displaystyle{C^n_r}$, we use the standard notation $\displaystyle{{n \choose r}}$. $\displaystyle{P(x \ge 2)}$ $\displaystyle{= 1 - P(x = 0) - P(x = 1)}$ $\displaystyle{= 1 - P(FFFF) - P(MFFF) - P(FMFF) - P(FFMF) - P(FFFM)}$ $\displaystyle{= 1 - P(FFFF) - {4 \choose 1} P(FFFM)}$

. $\displaystyle{P(FFFF) = \left( \frac{5}{11} \right) \left( \frac{4}{10} \right) \left( \frac{3}{9} \right) \left( \frac{2}{8} \right)}$ $\displaystyle{P(FFFM) = \left( \frac{5}{11} \right) \left( \frac{4}{10} \right) \left( \frac{3}{9} \right) \left( \frac{6}{8} \right)}$

. $\displaystyle{P(x \ge 2)}$ $\displaystyle{= \frac{6360}{7920}}$ $\displaystyle{= \frac{53}{66}}$

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Statistics method gets the final probability by dividing the number of ways of getting the desired result by the number of all possible results.

S method 1:

This method considers all the seats at once: what is the number of ways that they will get, for example, 3F and 1M?

In other words, instead of the seats, this method focuses on the people: what is the number of ways that 3F and 1M will be chosen?

In other words, instead of the choosing process, this method focuses on counting the results of the choosing.

Let $\displaystyle{N(...)}$ = the number of ways of … $\displaystyle{P(x \ge 2)}$ $\displaystyle{= \frac{N(\text{seats choose 2 or more men})}{\text{total number of ways}}}$ $\displaystyle{= \frac{\text{total number of ways} - N(\text{4 F among 5}) - N(\text{choose 3 F among 5 and then 1 M among 6})}{\text{total number of ways}}}$ $\displaystyle{= \frac{N(\text{seats choose any 4 people}) - N(\text{4 F among 5}) - N(\text{choose 3 F among 5}) \times N(\text{1 M among 6})}{N(\text{seats choose any 4 people})}}$ $\displaystyle{= \left[{11 \choose 4} - {5 \choose 4} - {5 \choose 3} {6 \choose 1} \right] / \left[ {11 \choose 4} \right]}$ $\displaystyle{= \frac{ 330 - 5 - 10 \times 6 }{330}}$ $\displaystyle{= \frac{ 53 }{ 66 }}$

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S method 2:

This method considers each seat one by one: what is the number of ways that this seat will get, for example, an M?

In other words, instead of the results, this method focuses on choosing process itself.

Note that this method uses permutation instead of combination. $\displaystyle{P(x \ge 2)}$ $\displaystyle{= \frac{\text{total number of ways} - N(FFFF) - N(MFFF) - N(FMFF) - N(FFMF) - N(FFFM)}{\text{total number of ways}}}$ $\displaystyle{= \frac{(11)(10)(9)(8) - (5)(4)(3)(2) - (6)(5)(4)(3) - (5)(6)(4)(3) - (5)(4)(6)(3) - (5)(4)(3)(6)}{(11)(10)(9)(8)}}$ $\displaystyle{= \frac{(11)(10)(9)(8) - (5)(4)(3)(2) - 4(6)(5)(4)(3)}{(11)(10)(9)(8)}}$ $\displaystyle{= \frac{6360}{7920}}$ $\displaystyle{= \frac{ 53 }{ 66 }}$

— Me@2021-10-02 05:18:02 PM

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