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I'm inviting 26 people over. Potential guests include 15 Bama fans, 10 Auburn fans, and 1 who doesn't like football. If only 10 come, what's the probability that all 10 are Bama fans?

And if 10 show up, what's the probability that 5 are Bama and 5 are Auburn?

jks
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2 Answers2

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There are $$\binom{15}{10}$$ ways to choose which Bama fans show up, so that is the number of successful outcomes.

The total number of outcomes is $$\binom{26}{10}$$

Therefore the probability is

$$\displaystyle\frac{\displaystyle\binom{15}{10}}{\displaystyle\binom{26}{10}} = \boxed{\frac{21}{37145}}$$

In the second problem, there are still $$\binom{26}{10}$$ total possibilities.

This time, there are $\binom{10}{5}$ ways to choose the Auburn fans and $\binom{15}{5}$ ways to choose the Bama fans, so there are

$$\binom{10}{5}\binom{15}{5}$$

successful outcomes.

This means that the probability is

$$\frac{\displaystyle\binom{10}{5}\displaystyle\binom{15}{5}}{\displaystyle\binom{26}{10}}=\boxed{\frac{5292}{37145}}$$

Zubin Mukerjee
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  • OK, starting to make sense...so what's the probability that the non-fan will show up? And given that as "x", is the chance that he won't show up 1-x? – jks Mar 13 '16 at 17:39
  • Are you asking "What's the probability that the non-fan will show up in a randomly selected set of 10 from the 26"? If so, it is the number of cases with the non-fan, which is $$\binom{25}{9}$$ divided by the total number of possibilities, which is still $$\binom{26}{10}$$ – Zubin Mukerjee Mar 13 '16 at 17:56
  • And yes the probability of something not happening is always one minus the probability of it happening. – Zubin Mukerjee Mar 13 '16 at 17:58
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To answer the first question, we first find how many total groups of $10$ people there are. This is simple - you invite $26$ people, so the number of groups of $10$ is $\dbinom{26}{10}.$ Now we count the number of groups with all $10$ people Bama fans. Since you invite $15$ Bama fans, there are $\dbinom{15}{10}$ groups of $10$ Bama fans. The probability is $\frac{\dbinom{15}{10}}{\dbinom{26}{10}} = \boxed{\frac{21}{37145}},$ which is approximately $0.0565\%.$

For the second case, our sample space is still $\dbinom{26}{10}.$ But now, we have to choose $5$ Bama fans and $5$ Auburn fans, which can be done in $\dbinom{15}{5} \times \dbinom{10}{5}$ ways. The probability is $\frac{\dbinom{15}{5} \times \dbinom{10}{5}}{\dbinom{26}{10}} = \boxed{\frac{5292}{37145}},$ which is about $14.2\%.$ This is much more likely than the first situation.

K. Jiang
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