# Setting Up Simple Probability Problem [duplicate]

This question already has an answer here:

I have attempted to set these up but want to make sure I'm on the right track. Any help is appreciated!

Dante can only invite 8 people to his birthday, but has 13 friends he’d like to invite. Eight of his friends are boys and five are girls. If Dante randomly selects 8 people, determine the probability for each scenario.

1. Dante does not invite any girls
$\binom{13}{8}$ Randomly selecting 8 people
$\binom{8}{8}$ not girls
= $\binom{13}{8}$ $\binom{8}{8}$

2. Dante invites an equal number of boys and girls
This would mean out of the 8 people, 4 need to be boys and 4 girls.
$\binom{13}{8}$ Selection
$\binom{8}{4}$ boys
$\binom{5}{4}$ girls
= $\binom{13}{8}$ ? $\binom{8}{4}$ * $\binom{5}{4}$

## marked as duplicate by lulu, Parcly Taxel, Joel Reyes Noche, hardmath, user223391 Oct 28 '16 at 16:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Note that a probability is a value between zero and one. You are counting outcomes, which is a good start, but the probability is obtained as a ratio of the favorable outcomes to all possible outcomes. – hardmath Oct 28 '16 at 14:45

## 1 Answer

Basically, you need to put the total number of possibilities in denumerator and the number of desired possibilities in the numerator.

1. $$\frac{\binom{8}{8}}{\binom{13}{8}}$$

2. $$\frac{\binom{8}{4}\binom{5}{4}}{\binom{13}{8}}$$