Definition of Probability

Now that we have the language to refer to outcomes and events of an experiment, we want to start quantifying how likely those outcomes/events are to occur.

A probability distribution on a sample space \Omega assigns probabilities to every event, satisfying the following conditions:

  1. \Pr(\Omega) = 1.

  2. 0 \leq \Pr(A) \leq 1 for any event A.

  3. If A \cap B = \emptyset, then \Pr(A\cup B) = \Pr(A) + \Pr(B).

For small probability spaces (i.e., with finitely many outcomes in the sample space), we'll usually assign probabilities to each individual outcome, and perhaps list them in a table. Then, to find the probability of any event, simply add together the probabilities of each outcome in that event.

An experiment consists of rolling a standard 6-sided die. The sample space is \Omega = \{1, 2, 3, 4, 5, 6\}. The probability distribution (assuming a fair die) is shown below.

Distribution for a fair die x \Pr(x) 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6

One possible event is A = \{2, 4, 6\}, i.e., the event that the result of the roll is even. The probability of A is: \Pr(A) = \Pr(2) + \Pr(4) + \Pr(6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}

Let \Omega = \{x_1, x_2, \dotsc, x_n\} be a sample space and A\subset \Omega an event. We refer to the distribution in which \Pr(x_i) = \frac{1}{n} for all i as the uniform distribution. In this case, it follows that \Pr(A) = \frac{|A|}{|\Omega|}.

When we talk about a fair coin flip or a fair die roll, the word "fair" is indicating a uniform distribution. However, don't make the mistake of assuming that all distributions are uniform by default.

A person picks a random number from 1 to 10. What is the probability that they picked 3?

Without assuming the distribution is fair (i.e., that each value 1, 2, \dotsc, 10 ) has probability 1/10 of occurring), we don't have enough information to answer this question.

In fact, the situation is even more vague than that: the sample space itself is unclear. Are we only allowed to pick integer values? What about fractions like 7/2? What about irrational numbers like \pi?

Consider the sample space \Omega = \{1, 2, 3, 4, 5, 6, 7, 8\} with probability distribution below. Calculate the probabilities of A = \{1, 3, 7, 8\}, B = \{2, 3, 6, 7\}, A\cup B, and A \cap B.

<tabular halign="center"> <row bottom="minor"> <cell><m>x</m></cell> <cell><m>\Pr(x)</m></cell> </row> <row> <cell>1</cell> <cell>0.1</cell> </row> <row> <cell>2</cell> <cell>0.05</cell> </row> <row> <cell>3</cell> <cell>0.2</cell> </row> <row> <cell>4</cell> <cell>0.15</cell> </row> <row> <cell>5</cell> <cell>0.15</cell> </row> <row> <cell>6</cell> <cell>0.1</cell> </row> <row> <cell>7</cell> <cell>0.05</cell> </row> <row> <cell>8</cell> <cell>0.1</cell> </row> <row> <cell>9</cell> <cell>0.1</cell> </row> </tabular> </table> </statement> <hint> <p> Remember that, to calculate the probability of an event, you should add up the probabilities of each outcome in the event. </p> </hint> <answer> <p> <m>\Pr(A) = 0.45, \Pr(B) = 0.4, \Pr(A \cup B) = 0.6, \Pr(A \cap B) = 0.25.</m> </p> </answer> <solution> <p> <md> <mrow> \Pr(A) \amp = \Pr(1) + \Pr(3) + \Pr(7) + \Pr(8) </mrow> <mrow> \amp = 0.1 + 0.2 + 0.05 + 0.1 </mrow> <mrow> \amp = 0.45 </mrow> <mrow> \Pr(B) \amp = \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) </mrow> <mrow> \amp = 0.05 + 0.2 + 0.1 + 0.05 </mrow> <mrow> \amp = 0.4 </mrow> </md> The other events are <m>A\cup B = \{1, 2, 3, 6, 7, 8\}</m> and <m>A\cap B = \{3, 7\}</m>, so: <md> <mrow> \Pr(A \cup B) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) + \Pr(8) </mrow> <mrow> \amp = 0.1 + 0.05 + 0.2 + 0.1 + 0.05 + 0.1 </mrow> <mrow> \amp = 0.6 </mrow> <mrow> \Pr(A \cap B) \amp = \Pr(3) + \Pr(7) </mrow> <mrow> \amp = 0.2 + 0.05 </mrow> <mrow> \amp = 0.25 </mrow> </md> </p> </solution> </exercise> <exercise> <introduction> <p> Suppose we flip a coin two times. Answer the questions below. What about three flips? What about four flips? </p> </introduction> <task> <statement> <p> Write all outcomes in the sample space <m>\Omega</m>. </p> </statement> <answer> <p> <m>\Omega = \{HH, HT, TH, TT\}.</m> </p> </answer> </task> <task> <statement> <p> Make a probability distribution table for <m>\Omega</m> assuming the coin is fair. </p> </statement> <answer> <table> <title>Probability Distribution for Two Fair Coin Flips x \Pr(x) HH 0.25 HT 0.25 TH 0.25 TT 0.25

Make a probability distribution table assuming the coin comes up heads with probability 0.3.

Probability Distribution for Two Fair Coin Flips x \Pr(x) HH 0.09 HT 0.21 TH 0.21 TT 0.49

Suppose we roll a fair 6-sided die two times. Answer the questions below.

Write all outcomes in the sample space \Omega.

\Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}

Make a probability distribution table for \Omega assuming the die is fair.

We'll avoid an overly large table and note that, since the die is fair, every outcome is equally likely. Therefore, \Pr(x) = \frac{1}{36} for every x\in \Omega.

Let A be the event that the second roll is higher than the first, and let B be the event that the first roll is even. Find \Pr(A), \Pr(B), and \Pr(A \cap B).

\Pr(A) = \frac{5}{12}, \Pr(B) = \frac{1}{2}, \Pr(A \cap B) = \frac{1}{6}.

A = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 4), (3, 5), (3, 6), \amp (4, 5), (4, 6), \amp (5, 6)\} B = \{ \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} A \cap B = \{ \amp (2, 3), (2, 4), (2, 5), (2, 6), \amp (4, 5), (4, 6)\} Therefore \Pr(A) = \frac{15}{36} = \frac{5}{12}, \Pr(B) = \frac{18}{36} = \frac{1}{2}, \Pr(A \cap B) = \frac{6}{36} = \frac{1}{6}.

Suppose a die has the values 1, 2, 3, 4, 5, 6 on the faces, but the die is not fair. Instead, the probabilities scale by the same amount as the face values. For example, a result of 4 is twice as likely as a result of 2, since 4 is twice as large as 2; a result of 6 is six times more likely than a result of 1; and so on. Write a probability distribution table for this die.

Probability Distribution for a Linearly Scaled Die x \Pr(x) 1 1/21 2 2/21 3 3/21 4 4/21 5 5/21 6 6/21

Let \Pr(1) = x. Then \Pr(2) = 2 \Pr(1) = 2x, and \Pr(3) = 3\Pr(1) = 3x, and so on. So the total probability in the space is: Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) \amp = x + 2x + 3x + 4x + 5x + 6x \amp = 21x Since the total probability must add up to 1, we have 1 = 21x, so x = \frac{1}{21}, and we can calculate the rest of the probabilities from there.

Probability Distribution for a Linearly Scaled Die x \Pr(x) 1 1/21 2 2/21 3 3/21 4 4/21 5 5/21 6 6/21

Suppose a die has the values 1, 2, 3, 4, 5, 6 on the faces, but the die is not fair. Instead, each even value has an equal probability, each odd value has an equal probability, and the even values are each twice as likely as the odd values to appear on a roll. Write a probability distribution table for this die.

Probability Distribution for an Even-biased Die x \Pr(x) 1 1/9 2 2/9 3 1/9 4 2/9 5 1/9 6 2/9

Let \Pr(1) = x. Then \Pr(3) = x and \Pr(5) = x, since all odd rolls must have the same probability. The even rolls must have twice the probability, so \Pr(2) = \Pr(4) = \Pr(6) = 2x. Now: Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) \amp = x + 2x + x + 2x + x + 2x \amp = 9x Since the total probability must add up to 1, we have 1 = 9x, so x = \frac{1}{9}, and we can calculate the rest of the probabilities from there.

Probability Distribution for an Even-biased Die x \Pr(x) 1 1/9 2 2/9 3 1/9 4 2/9 5 1/9 6 2/9

A toxin molecule inside a cell has a 0.3 probability of leaving the cell during a 1-minute period. For each value of n = 1, 2, 3, \dotsc, find the probability of the toxin molecule leaving the cell during the n th minute. What is the probability of the molecule leaving the cell during the first 3 minutes?

Write T for the minute that the toxin molecule leaves the cell. Then \Pr(T = n) = (0.7)^{n - 1} (0.3), and \Pr(T \leq 3) = 0.657.

Write T for the minute that the toxin molecule leaves the cell. We're told that the toxin molecule has probability 0.3 of leaving during each minute. Therefore, the probability of remaining during a particular minute is 1 - 0.3 = 0.7.

For the toxin molecule to leave the cell during minute n, it must remain for minutes 1, 2, \dotsc, n - 1, and then leave during minute n. The probability to remain each minute is 0.7, so we must multiply n - 1 copies of 0.7. Then the probability of leaving is 0.3, so we multiply by a factor of 0.3, yielding the formula: \Pr(T = n) = (0.7)^{n - 1}(0.3).

Then, the probaiblity of the molecule leaving within the first three minutes will be: \Pr(T \leq 3) \amp = \Pr(T = 1) + \Pr(T = 2) + \Pr(T = 3) \amp = 0.3 + (0.7)(0.3) + (0.7)^2(0.3) \amp = 0.657.