Equally likely outcomes
Theorem taught in Random experiments and probability, part of Probability
Equally likely outcomes
| Type | Theorem |
| Taught in | Random experiments and probability |
| Part | Probability |
| Worked examples | 3 |
If \Omega is finite and every outcome has the same probability, then for every event A
\mathbb{P}(A)=\frac{|A|}{|\Omega|}.
Examples
The birthday problem
Worked example from Random experiments and probability
A group has n people. Ignore leap days and assume each birthday is equally likely to be any of 365 days, independently.
Question. What is the probability that at least two people share a birthday?
Solution. The birthday count is the second row of the table against the first. The probability that all birthdays differ is
\frac{365\times 364\times\cdots\times(365-n+1)}{365^n},
provided n\le 365, and zero otherwise. The numerator is the number of injective assignments of days, and the denominator is the number of assignments with replacement. The probability of a shared birthday is one minus this. It passes one half at n=23, where it is 0.507, and reaches 0.891 at n=40 and 0.970 at n=50.
The surprise comes from counting the wrong thing. 23 people seem few against 365 days, but they form \binom{23}{2}=253 pairs, and any one of those pairs can match.
Matching faces
Question page: Matching faces
Two fair dice are thrown, and the ordered pair is recorded.
Find the probability that the two faces agree.
Find the same probability given that the sum of the faces is at most 4.
The first face may be anything. The second matches it with probability 1/6. There are 6 matching pairs out of 36.
Let A be the event that the faces agree, and let B be the event that the sum is at most 4. Then
B=\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\}, \qquad A\cap B=\{(1,1),(2,2)\}.
Every ordered pair is equally likely, so \mathbb{P}(A\mid B)=2/6=1/3.
One of each court card
Question page: One of each court card
Four cards are drawn at once from a deck of 52, so the hand is an unordered set. Find the probability that the hand consists of one king, one queen, one jack and one ace.
The sample space has \binom{52}{4} equally likely hands. The favourable hands choose one card from each of the four ranks, so there are 4^4 of them. The probability is 4^4/\binom{52}{4}.
Used in
- Taught in Random experiments and probability, where the derivation and examples are.
- All 17 questions on Random experiments and probability
See also
Sample space · Event · Probability measure · Properties of a probability · Four ways of counting draws · Conditional probability · Product rule · Law of total probability