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Combinatorics and Probability

Probability and counting

Let \(\Omega\) be a finite sample space in which every outcome is equally likely, and let \(A\) be an event. Then \[P(A) = \frac{|A|}{|\Omega|}\] where \(|A|\) is the number of elements in \(A\) and \(|\Omega|\) is the number of elements in \(\Omega.\)

1. There are six balls in an urn. Four of the balls are red and the other two are blue. If a ball is chosen at random, what is the probability it is red?

▼ Answer:
\(\frac{2}{3}\)

▼ Solution:
There are \(6\) balls in the urn, so the size of the sample space is \(|\Omega| = 6.\) The event \(A\) in question is about the probability of drawing a red ball. There are \(4\) red balls, so \(|A| = 4.\) Therefore, the probability of drawing a red ball is \[P(A) = \frac{4}{6} = \frac{2}{3}\]

2. Example: If you roll a \(6\) sided die, what the the probability you get a \(4\) or higher?

▼ Answer:
\(\frac{1}{2}\)

▼ Solution:
There are \(6\) possible outcomes for the die roll, so the size of the sample space is \(|\Omega| = 6.\) The event \(A\) in question is about the probability of rolling a \(4\) or higher. The numbers on a die that are \(4\) or higher are \(4, 5, 6\), so \(|A| = 3.\) Therefore, the probability of rolling a \(4\) or higher is \[P(A) = \frac{3}{6} = \frac{1}{2}\]

3. There are \(3\) races. The first race has \(10\) competitors, \(2\) of which are from the U.S.A. The second race has \(9\) competitors, \(3\) of which are from the U.S.A. The third race has \(12\) competitors, \(4\) of which are from the U.S.A. If every competitor is equally likely to win each race, what is the probability that the winners of all \(3\) races are from the U.S.A.?

▼ Answer:
\(\frac{1}{45}\)

▼ Solution 1 (Counting):

▼ Solution 2 (Independent Events):