Lecture 12 – Probability

DSC 10, Fall 2022

Announcements

• Lab 4 is due Saturday at 11:59PM.
• Homework 4 is due on Tuesday 10/25 at 11:59PM.
• The Midterm Project will be released today and is due Tuesday 11/1.
  • It takes much longer than a homework, so start now!
  • Partners are optional but recommended, and can be from any lecture section.
  • If you're looking for a partner, email tutor Anna (a2liu@ucsd.edu).
  • You must use the pair programming model when working with a partner.

Agenda

We'll cover the basics of probability theory. This is a math lesson; take written notes. ✍🏽

Probability resources

Probability is a tricky subject. If it doesn't click during lecture or on the assignments, take a look at the following resources:

• Computational and Inferential Thinking, Chapter 9.5.
• Theory Meets Data, Chapters 1 and 2.
• Khan Academy's unit on Probability.

Probability theory

Probability theory

• Some things in life seem random.
  • e.g. flipping a coin or rolling a die 🎲.
• The probability of seeing "heads" when flipping a fair coin is $\frac{1}{2}$.
• One interpretation of probability says that if we flipped a coin infinitely many times, then $\frac{1}{2}$ of the outcomes would be heads.

Terminology

• Experiment: A process or action whose result is random.
  • e.g., rolling a die.
  • e.g., flipping a coin twice.
• Outcome: The result of an experiment.
  • e.g., the possible outcomes of rolling a six-sided die are 1, 2, 3, 4, 5, and 6.
  • e.g., the possible outcomes of flipping a coin twice are HH, HT, TH, and TT.
• Event: A set of outcomes.
  • e.g., the event that the die lands on a even number is the set of outcomes {2, 4, 6}.
  • e.g., the event that the die lands on a 5 is the set of outcomes {5}.
  • e.g., the event that there is at least 1 head in 2 flips is the set of outcomes {HH, HT, TH}.

Terminology

• Probability: A number between 0 and 1 (equivalently, between 0% and 100%) which describes the likelihood of an event.
  • 0: the event never happens.
  • 1: the event always happens.
• Notation: if $A$ is an event, $P(A)$ is the probability of that event.

Equally-likely outcomes

• If all outcomes in event $A$ are equally likely, then the probability of $A$ is

$$ P(A) = \frac{ \text{# of outcomes satisfying $A$} }{ \text{total # of outcomes} } $$

• Example 1: Suppose we flip a fair coin 3 times. What is the probability we see exactly 2 heads?

Example 1 solved

• When we flip a fair coin 3 times, there are 8 possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT.
• These outcomes are all equally likely.
• 3 of these outcomes have exactly 2 heads: HHT, HTH, and THH.
• So, the probability of seeing exactly 2 heads in 3 flips of a fair coin is $\frac{3}{8}$.

Concept Check ✅ – Answer at cc.dsc10.com

I have three cards: red, blue, and green. What is the chance that I choose a card at random and it is green, then – without putting it back – I choose another card at random and it is red?

• A) $\frac{1}{9}$
• B) $\frac{1}{6}$
• C) $\frac{1}{3}$
• D) $\frac{2}{3}$
• E) None of the above.

Solution

• There are 6 possible outcomes: RG, RB, GR, GB, BR, and BG.
• These outcomes are equally likely.
• There is only 1 outcome which makes the event happen: GR.
• Hence the probability is $\frac{1}{6}$.

Conditional probabilities

• Two events $A$ and $B$ can both happen. Suppose that we know $A$ has happened, but we don't know if $B$ has.
• If all outcomes are equally likely, then the conditional probability of $B$ given $A$ is:

$$ P(B \text{ given } A) = \frac{ \text{# of outcomes satisfying both $A$ and $B$} }{ \text{# of outcomes satisfying $A$} } $$

• Intuitively, this is similar to the definition of the regular probability of $B$, $P(B) = \frac{ \text{# of outcomes satisfying $B$} }{ \text{total # of outcomes} }$, if you restrict the set of possible outcomes to be just those in event $A$.

Concept Check ✅ – Answer at cc.dsc10.com

$$ P(B \text{ given } A) = \frac{ \text{# of outcomes satisfying both $A$ and $B$} }{ \text{# of outcomes satisfying $A$} } $$

I roll a six-sided die and don't tell you what the result is, but I tell you that it is 3 or less. What is the probability that the result is even?

• A) $\frac{1}{2}$
• B) $\frac{1}{3}$
• C) $\frac{1}{4}$
• D) None of the above.

Solution

$$ P(B \text{ given } A) = \frac{ \text{# of outcomes satisfying both $A$ and $B$} }{ \text{# of outcomes satisfying $A$} } $$

• $A$ is the event "roll is 3 or less", $B$ is the event "roll is even".
• There are three outcomes where the roll is 3 or less: 1, 2, and 3.
• There is only one outcome where the roll is 3 or less and even: 2.
• So the probability that the roll is even given that it is 3 or less is $P(B \text{ given } A) = \frac{1}{3}$.

Probability that two events both happen

• Suppose again that $A$ and $B$ are two events, and that all outcomes are equally likely. Then, the probability that both $A$ and $B$ occur is

$$ P(A \text{ and } B) = \frac{ \text{# of outcomes satisfying both $A$ and $B$} }{ \text{total # of outcomes} } $$

• Example 2: I roll a fair six-sided die. What is the probability that the roll is 3 or less and even?

Example 2 solved

I roll a fair six-sided die. What is the probability that the roll is 3 or less and even?

• Only one outcome is both 3 or less and even: 2.
• There are 6 total outcomes.
• Thus, $P(A \text{ and } B) = \frac{1}{6}$.

The multiplication rule

• The multiplication rule specifies how to compute the probability of both $A$ and $B$ happening, even if all outcomes are not equally likely.

$$ P(A \text{ and } B) = P(A) \cdot P(B \text{ given } A) $$

• Example 2, again: I roll a fair six-sided die. What is the probability that the roll is 3 or less and even?

Example 2 solved, again

I roll a fair six-sided die. What is the probability that the roll is 3 or less and even?

• The probability that the roll is 3 or less is $P(A) = \frac{1}{2}$.
• From before, the probability that the roll is even given that the roll is 3 or less is $P(B \text{ given } A) = \frac{1}{3}$.
• Thus, the probability the roll is both 3 or less and even is $P(A \text{ and } B) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$.
• Note that an equivalent formula is $P(A \text{ and } B) = P(B) \cdot P (A \text{ given } B)$.

Generally, situations involving an "and" involve multiplication.

What if $A$ isn't affected by $B$? 🤔

• The multiplication rule states that, for any two events $A$ and $B$, $$P(A \text{ and } B) = P(A) \cdot P(B \text{ given } A)$$
• What if knowing that $A$ happens doesn't tell you anything about the likelihood of $B$ happening?
  • Suppose we flip a fair coin three times.
  • The probability that the second flip is heads doesn't depend on the result of the first flip.
• Then, what is $P(A \text{ and } B)$?

Independent events

• Two events $A$ and $B$ are independent if $P(B \text{ given } A) = P(B) \:$, or equivalently if $$P(A \text{ and } B) = P(A) \cdot P(B)$$
• Example 3: Suppose we have a coin that is biased, and flips heads with probability 0.7. Each flip is independent of all other flips. We flip it 5 times. What's the probability we see 5 heads in a row?

Example 3 solved

Suppose we have a coin that is biased, and flips heads with probability 0.7. Each flip is independent of all other flips. We flip it 5 times. What's the probability we see 5 heads in a row?

• The probability of seeing heads on a single flip is 0.7.
• Each flip is independent.
• So, the probability of seeing 5 heads in a row is

$$0.7 \cdot 0.7 \cdot 0.7 \cdot 0.7 \cdot 0.7 = 0.7^5$$

Probability that an event doesn't happen

• The probability that $A$ doesn't happen is $1 - P(A) \:$.
• For example, if the probability it is sunny tomorrow is 0.85, then the probability it is not sunny tomorrow is 0.15.

Concept Check ✅ – Answer at cc.dsc10.com

Every time I call my grandma 👵, the probability that she answers her phone is $\frac{1}{3}$. If I call my grandma three times today, what is the chance that I will talk to her at least once?

• A) $\frac{1}{3}$
• B) $\frac{2}{3}$
• C) $\frac{1}{2}$
• D) $1$
• E) None of the above.

Solution

• Let's first calculate the probability that she doesn't answer her phone in three tries.
  • The probability she doesn't answer her phone on any one attempt is $\frac{2}{3}$.
  • So the probability she doesn't answer her phone in three tries is $\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{27}$.
• But we want the probability of her answering at least once. So we subtract the above result from 1.
  • $1 - \frac{8}{27} = \frac{19}{27}$; none of the above!

Probability of either of two events happening

• Suppose again that $A$ and $B$ are two events, and that all outcomes are equally likely. Then, the probability that either $A$ or $B$ occur is

$$ P(A \text{ or } B) = \frac{ \text{# of outcomes satisfying either $A$ or $B$} }{ \text{total # of outcomes} } $$

• Example 4: I roll a fair six-sided die. What is the probability that the roll is even or more than 5?

Example 4 solved

I roll a fair six-sided die. What is the probability that the roll is even or at least 5?

• There are three outcomes that are even: 2, 4, 6.
• There are two outcomes that are at least 5: 5, 6.
• There are four total outcomes that satisfy at least one of the two conditions: 2, 4, 5, 6.
• Thus, the probability that the roll is even or at least 5 is $\frac{4}{6} = \frac{2}{3}$.
  • Note that this is not $P(A) + P(B) \:$, which would be $\frac{3}{6} + \frac{2}{6} = \frac{5}{6}$, because there is overlap between events $A$ and $B$.

The addition rule

• Suppose that if $A$ happens, then $B$ doesn't, and if $B$ happens, then $A$ doesn't.
  • Such events are called mutually exclusive – they have no overlap.
• If $A$ and $B$ are any two mutually exclusive events, then

$$P(A \text{ or } B) = P(A) + P(B)$$

• Example 5: Suppose I have two biased coins, coin $A$ and coin $B$. Coin $A$ flips heads with probability 0.6, and coin $B$ flips heads with probability 0.3. I flip both coins once. What's the probability I see two different faces?

Example 5 solved

Suppose I have two biased coins, coin $A$ and coin $B$. Coin $A$ flips heads with probability 0.6, and coin $B$ flips heads with probability 0.3. The two coins are independent of one another. I flip both coins once. What's the probability I see two different faces?

• The event we see two different faces corresponds to either seeing a head then a tail, or a tail then a head (i.e. not both heads and not both tails).
• The probability of seeing a head then a tail is $0.6 \cdot (1 - 0.3)$, because the two coins are independent of one another.
• The probability of seeing a tail then a head is $(1 - 0.6) \cdot 0.3$.
• So, the probability of seeing two different faces is

$$0.6 \cdot (1 - 0.3) + (1 - 0.6) \cdot 0.3 = 0.54$$

Generally, situations involving an "or" involve addition.

Aside: proof of the addition rule for equally-likely events

You are not required to know how to "prove" anything in this course; you may just find this interesting.

If $A$ and $B$ are events consisting of equally likely outcomes, and furthermore $A$ and $B$ are mutually exclusive (meaning they have no overlap), then

$$ \begin{align*} P(A \text{ or } B) &= \frac{ \text{# of outcomes satisfying either $A$ or $B$} }{ \text{total # of outcomes} } \\[1em] &= \frac{ (\text{# of outcomes satisfying $A$}) + (\text{# of outcomes satisfying $B$}) }{ \text{total # of outcomes} } \\[1em] &= \frac{ (\text{# of outcomes satisfying $A$}) }{ \text{total # of outcomes} } + \frac{ (\text{# of outcomes satisfying $B$}) }{ \text{total # of outcomes} } \\[1em] &= P(A) + P(B) \end{align*} $$

Summary

Summary

• Probability describes the likelihood of an event occurring.
• There are several rules for computing probabilities. We looked at many special cases that involved equally-likely events.
• There are two general rules to be aware of:
  • The multiplication rule, which states that for any two events, $P(A \text{ and } B) = P(B \text{ given } A) \cdot P(A) \:$.
  • The addition rule, which states that for any two mutually exclusive events, $P(A \text{ or } B) = P(A) + P(B)$.
• Next time: simulations.