Probability Scale: \(0 \le \operatorname{P}(A) \le 1\) for any event \(A\).
Total Probability: \(\operatorname{P}(S) = 1\) where \(S\) is the sample space.
Additivity: If \(A\) and \(B\) are mutually exclusive events, then \[\operatorname{P}(A \operatorname{or}B) = \operatorname{P}(A) + \operatorname{P}(B) \;.\]
Equally Likely Rule: If all outcomes are equally likely, then \[ \operatorname{P}(A) = \frac{\mbox{number of outcomes in A}}{\mbox{number of outcomes in sample space}} \;. \]
Complement Rule: \(\operatorname{P}(A^c) = 1 - \operatorname{P}(A)\)
General Addition Rule: \(\operatorname{P}(A \operatorname{or}B) = \operatorname{P}(A) + \operatorname{P}(B) - \operatorname{P}(A \operatorname{and}B)\)
Multiplication Rule: \(\operatorname{P}(A \operatorname{and}B) =\)
Notation: \(\operatorname{P}(A \operatorname{if}B)\) or \(\operatorname{P}(A \mid B)\).
We will read this one of these ways:
Asking the question: What proportion of the time that B happens does A also happen.
Definition:
\[\operatorname{P}(A \operatorname{if}B) = \frac{\operatorname{P}(A \operatorname{and}B)}{\operatorname{P}(B)} = \frac{\operatorname{P}(\mbox{both})}{\operatorname{P}(\mbox{condition})} \]
If we roll a fair die (6 sides numbered 1 through 6, each equally likely to be rolled), what is
Cystic fibrosis (CF) is a life-threatening genetic disorder caused by mutations in the CFTR gene located on chromosome 7. Defective copies of CFTR can result in the reduced quantity and function of the CFTR protein, which leads to the buildup of thick mucus in the lungs and pancreas. CF is an autosomal recessive disorder; an individual only develops CF if they have inherited two affected copies of CFTR (one from each parent). Individuals with one normal (wild-type) copy and one defective (mutated) copy are known as carriers; they do not develop CF, but may pass the disease-causing mutation onto their offspring.
A standard deck of cards has 52 cards. There are four suits (clubs, diamonds, hearts, spades) and 13 denominations in each suit (Ace, 2, 3, 4, 5, 6, 7, 8, 9, jack, queen, king). Clubs and spades are black. Diamonds are hearts are red.
If you are dealt 1 card from a standard deck, what is
If we select one card from a shuffled deck of cards, what is the probability that the card
T-shirts. A survey of a 5th grade students at a school asked them whether they preferred Red or Yellow for a class T-shirt and also recorded whether each child was a boy or a girl. Let \(B\) be the event that a child is a boy, \(G\) that the child is a girl, \(R\) that the child prefers red, and \(Y\) that the child prefers yellow. Suppose we randomly select one survey respondent from the class. Compute each of these probabilities using the table below.
| . | Red | Yellow |
|---|---|---|
| Boy | 17 | 8 |
| Girl | 12 | 16 |
\(\operatorname{P}(B)\)
\(\operatorname{P}(G)\)
\(\operatorname{P}(R)\)
\(\operatorname{P}(B \operatorname{and}R)\)
\(\operatorname{P}(R \operatorname{and}B)\)
\(\operatorname{P}(B \operatorname{or}R)\)
\(\operatorname{P}(R \operatorname{or}B)\)
\(\operatorname{P}(B \operatorname{if}R)\)
\(\operatorname{P}(R \operatorname{if}B)\)
\(\operatorname{P}(Y \operatorname{if}B)\)
\(\operatorname{P}(B \operatorname{if}Y)\)
The multiplication rule.
Independence. Sometimes \(\operatorname{P}(A \operatorname{if}B) = \operatorname{P}(A)\). That means knowing that \(B\) happens doesn’t change our probability for \(A\) – it adds no information useful for computing the probability that \(A\) happens. When this happens, we say that \(A\) and \(B\) are independent events.1
What is the Multiplication Rule for Independent Events?
Toss a coin twice. Let \(H_1\) be the event that the first toss is a head. Let \(H_2\) be the even that the second toss is a head.
Shuffle a deck of cards and deal two of them. Let \(B_1\) be the event that the first card is black. Let \(B_2\) be the event that the second card is black.
Roll two standard dice. Let \(A\) be the event that the first die is even. Let \(B\) be the event that the second die is a six.
Can mutually exclusive events be independent? Give an example or explain why not.
Hospital. Suppose that 35% of all patients admitted to a hospital’s intensive care unit have high blood pressure, 42% have some sort of infection, and 12% have both problems.
If we draw two cards from a deck of cards, what is the probability that
Flip a coin 4 times, what is the probability of getting heads each time? What if you flip it 10 times?
Suppose you conduct a hypothesis test with \(\alpha = 0.05\) and the null hypothesis is true.
In a bowl are 4 red marbles and 2 blue marbles. If you reach in without looking and select two of the marbles, let \(X\) be the number of blue marbles. Fill in the following probability table.
| value of X | 0 | 1 | 2 |
|---|---|---|---|
| probability |
Do your probabilities add up to 1? Should they?
Are the events \(X=0\), \(X=1\), and \(X=2\) equally likely?
T-shirts
Multiplication Rule:
If \(A\) and \(B\) are independent, then \(\operatorname{P}(A \operatorname{and}B) = \operatorname{P}(A) \cdot \operatorname{P}(B)\). But this only holds if \(A\) and \(B\) are independent.
Hospital
Cards
Coin
Hypothesis Test
Bowl of marbles
It always goes both ways. If \(\operatorname{P}(A \operatorname{if}B) = \operatorname{P}(A)\), it will also be the case that \(\operatorname{P}(B \operatorname{if}A) = \operatorname{P}(B)\). If you like to do a little algebra, you can show that this is so.↩︎