Before you start, make sure you know: Concepts (Math 7)
Before you start, make sure you know: Probability (Math 7)
Your teacher pulls names from a hat to pick two students for a demo, and she doesn't put the first name back. Once Ashley's name is out, every other name has a slightly better chance of being picked next. A doctor reads a test result the same way: what she already knows (a positive test) changes the probability of what she wants to find out (whether you are sick). In both cases one event changes the probability of the other. These are dependent events, and you need conditional probability to handle them.
Conditional Probability
Why does this work? Once you know $A$ happened, $A$ becomes your whole sample space. Only the part of $B$ that lies inside $A$ still counts, and you measure it against $A$, not against everything.
A two-way table makes this easy to see. Thirty students were asked whether they play a sport and whether they play an instrument.
Sport
No sport
Total
Instrument
7
5
12
No instrument
13
5
18
Total
20
10
30
The same idea works with dice. Here is a grid of all 36 outcomes (first die down, second die across). The highlighted cells are the outcomes with sum 8.
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In a class of $30$ students, $18$ study French, $12$ study German and $5$ study both languages. If a student is chosen at random and it turns out they study French, what…
In a class of $30$ students, $18$ study French, $12$ study German and $5$ study both languages. If a student is chosen at random and it turns out they study French, what is the probability that they also study German?
We roll two normal six-sided dice. Knowing that the sum of the scores obtained is $8$, what is the probability that a $3$ has been rolled on one of the dice?
We roll two normal six-sided dice. Knowing that the sum of the scores obtained is $8$, what is the probability that a $3$ has been rolled on one of the dice?
The Multiplication Rule and Drawing Without Replacement
Why? For $A$ and then $B$ to happen, $A$ must happen first. Then, in the new situation $A$ left behind, $B$ must happen with its updated probability.
Think of the favorites as red balls and the other songs as blue balls. Draw two and watch the counts change.
The conditional probability does not have to come from counting. Sometimes it is given as a percent.
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In an urn there are $5$ red balls and $3$ blue balls. Two balls are drawn consecutively and without replacement. Calculate the probability that the first ball is red and…
In an urn there are $5$ red balls and $3$ blue balls. Two balls are drawn consecutively and without replacement. Calculate the probability that the first ball is red and the second is blue.
From a Spanish deck of $40$ cards, two cards are drawn at random one after the other and without replacement. What is the probability that both cards are kings?
From a Spanish deck of $40$ cards, two cards are drawn at random one after the other and without replacement. What is the probability that both cards are kings?
In a factory, machine A produces $60\%$ of the parts and machine B produces the remaining $40\%$. It is known that $2\%$ of the parts produced by A are defective, while…
In a factory, machine A produces $60\%$ of the parts and machine B produces the remaining $40\%$. It is known that $2\%$ of the parts produced by A are defective, while on machine B the percentage of defective parts is $5\%$. If a part is chosen at random from the warehouse, what is the probability that it was manufactured by machine B and is defective?
A tree diagram keeps a multi-stage experiment organized. Each stage is a set of branches. The probability on a branch is conditional on the branches before it. Here two marbles are drawn without replacement from a bag of 3 red (R) and 2 blue (B).
Sometimes the second stage is not just "one fewer object." It is a completely different situation that depends on the first stage.
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Observe the following tree diagram that represents a two-stage compound experiment. Calculate the probability of obtaining the final result $C$ passing through branch…
Observe the following tree diagram that represents a two-stage compound experiment. Calculate the probability of obtaining the final result $C$ passing through branch $A$.
We have two boxes. BOX I contains $4$ green balls and $2$ yellow balls. BOX II contains $3$ green balls and $5$ yellow balls. We draw a ball at random from BOX I and…
We have two boxes. BOX I contains $4$ green balls and $2$ yellow balls. BOX II contains $3$ green balls and $5$ yellow balls. We draw a ball at random from BOX I and place it into BOX II. Then, we mix well and draw a ball from BOX II. What is the probability that both drawn balls are green?
With three or more draws, the multiplication rule just keeps going:
$$P(A \text{ and } B \text{ and } C) = P(A) \cdot P(B \mid A) \cdot P(C \mid A \text{ and } B)$$
Each factor uses whatever is left after the draws before it.
Sometimes the unknown is how many objects there are. Then the probability becomes an equation.
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From a box containing $10$ light bulbs, it is known that $3$ of them are blown. $3$ light bulbs are drawn successively and without replacement. What is the probability…
From a box containing $10$ light bulbs, it is known that $3$ of them are blown. $3$ light bulbs are drawn successively and without replacement. What is the probability that exactly one of the drawn light bulbs is blown?
An opaque bag contains $10$ slips of paper identical to the touch: $2$ of them have a prize and $8$ are blank. Three friends, Ashley, Brandon, and Daniel, draw…
An opaque bag contains $10$ slips of paper identical to the touch: $2$ of them have a prize and $8$ are blank. Three friends, Ashley, Brandon, and Daniel, draw consecutively and without replacement one slip each in that same order. What is the probability that the first prize drawn is drawn by Daniel?
In an urn there are $x$ red balls and $4$ black balls. Two balls are drawn consecutively and without replacement. Knowing that the probability that both balls are red is…
In an urn there are $x$ red balls and $4$ black balls. Two balls are drawn consecutively and without replacement. Knowing that the probability that both balls are red is exactly $\frac{1}{6}$, set up an equation and determine the number of red balls $x$ that are in the urn.
Two events are independent if knowing that one happened does not change the probability of the other.
Why are these the same? The multiplication rule says $P(A \text{ and } B) = P(A) \cdot P(B \mid A)$. If $P(B \mid A)$ is just $P(B)$, the rule becomes $P(A) \cdot P(B)$.
To test independence, you sometimes first need $P(A \text{ and } B)$ from other information. That takes the addition rule.
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Consider two events $A$ and $B$ of a random experiment. We know that $P(A) = 0,4$, $P(B) = 0,5$ and $P(A \cap B) = 0,2$. Check mathematically if events $A$ and $B$ are…
Consider two events $A$ and $B$ of a random experiment. We know that $P(A) = 0,4$, $P(B) = 0,5$ and $P(A \cap B) = 0,2$. Check mathematically if events $A$ and $B$ are independent.
In a town, $45\%$ of the inhabitants read newspaper X, $35\%$ read newspaper Y, and $15\%$ read both newspapers. If we choose an inhabitant at random and we know that…
In a town, $45\%$ of the inhabitants read newspaper X, $35\%$ read newspaper Y, and $15\%$ read both newspapers. If we choose an inhabitant at random and we know that they read newspaper X, what is the probability that they also read newspaper Y? In light of the results, reason whether the events "reading newspaper X" and "reading newspaper Y" are independent.
Let $A$ and $B$ be two events of the same random experiment such that $P(A) = 0,5$, $P(B) = 0,4$ and $P(A \cup B) = 0,7$. Using the properties of probability, find the…
Let $A$ and $B$ be two events of the same random experiment such that $P(A) = 0,5$, $P(B) = 0,4$ and $P(A \cup B) = 0,7$. Using the properties of probability, find the value of the intersection and calculate $P(A|B)$ and $P(B|A)$. Are $A$ and $B$ independent events? Reason your answer.
When events are independent, the multiplication rule gets simpler: $P(A \text{ and } B) = P(A) \cdot P(B)$. Drawing with replacement gives independent stages. So do separate things that don't affect each other. If you draw a card and then flip a coin, the coin doesn't care what the card was. The chance of an ace and heads is $\frac{4}{52} \cdot \frac12 = \frac{1}{26}$.
"At least one": use the complement
"At least one" is usually easier through its opposite, "none."
$$P(\text{at least one}) = 1 - P(\text{none})$$
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A coin is biased so that the probability of getting Heads is $0,7$. The coin is flipped twice consecutively. Calculate the probability of getting exactly one Heads and…
A coin is biased so that the probability of getting Heads is $0,7$. The coin is flipped twice consecutively. Calculate the probability of getting exactly one Heads and one Tails (in any order).
Two archers shoot an arrow at the same target. Archer A hits with a probability of $0,7$ and archer B hits with a probability of $0,8$. Their shots are completely…
Two archers shoot an arrow at the same target. Archer A hits with a probability of $0,7$ and archer B hits with a probability of $0,8$. Their shots are completely independent. What is the probability that the target is hit by at least one arrow?
A computer security system has two independent firewalls. The probability that firewall A fails against an attack is $0,15$ and the probability that firewall B fails is…
A computer security system has two independent firewalls. The probability that firewall A fails against an attack is $0,15$ and the probability that firewall B fails is $0,10$. The system is compromised only if both firewalls fail. What is the probability that the system resists an attack? If a third firewall C, identical to B and independent of the previous ones, is added, by how much does the probability that the system resists the attack increase?
Sometimes an event can happen through several different "first stages." Jayden gets to school by bike on 70% of days and by bus on the other 30%. He is late 10% of the time when he bikes and 20% of the time when he takes the bus. How often is he late?
Why? Every path to $B$ goes through exactly one of the $A_i$. You find the probability of each path with the multiplication rule and add them.
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A basketball player has a success rate of $80\%$ on the first free throw. If they make the first one, their confidence increases and the probability of making the second…
A basketball player has a success rate of $80\%$ on the first free throw. If they make the first one, their confidence increases and the probability of making the second free throw goes up to $90\%$. However, if they miss the first one, they get nervous and the probability of making the second one drops to $60\%$. Calculate the probability that the player scores exactly one free throw in a series of two shots.
A rapid allergy test is performed in a clinic. It is known that $8\%$ of the patients are allergic to a certain substance. The test gives a positive result in $90\%$ of…
A rapid allergy test is performed in a clinic. It is known that $8\%$ of the patients are allergic to a certain substance. The test gives a positive result in $90\%$ of the patients who are truly allergic. However, it also gives a false positive in $5\%$ of the patients who are not allergic. A patient is chosen at random and the test is performed on them. What is the probability that the result is positive?
We have three chests that are identical on the outside. CHEST A contains $2$ gold coins and $1$ silver coins. CHEST B contains $1$ gold coin and $3$ silver coins. CHEST…
We have three chests that are identical on the outside. CHEST A contains $2$ gold coins and $1$ silver coins. CHEST B contains $1$ gold coin and $3$ silver coins. CHEST C contains $4$ gold coins. We roll a fair die: if it shows $1$ or $2$, we choose chest A; if it shows $3$, we choose chest B; if it shows $4$, $5$ or $6$, we choose chest C. Then, we draw a coin at random from the chosen chest. What is the probability that the drawn coin is gold?
Now flip the question. Jayden walks in late. What is the probability that he came by bus? You know $P(\text{late} \mid \text{bus})$, but you want $P(\text{bus} \mid \text{late})$.
Why? This is just the definition $\dfrac{P(A \text{ and } B)}{P(B)}$. The numerator is the one path you care about, and the denominator is all the paths to $B$.
A "natural frequency" picture can help. Imagine 1,000 school days. Jayden is late on 70 bike days and 60 bus days, so 130 late days in all.
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On a bookshelf there are $5$ novels and $3$ poetry books. Two books are chosen at random one after the other and without replacement. If the second chosen book turned…
On a bookshelf there are $5$ novels and $3$ poetry books. Two books are chosen at random one after the other and without replacement. If the second chosen book turned out to be poetry, what is the probability that the first chosen book was too?
In a high school, $55\%$ of the students are girls. It is known that $20\%$ of the girls and $25\%$ of the boys actively participate in the drama club. If a student who…
In a high school, $55\%$ of the students are girls. It is known that $20\%$ of the girls and $25\%$ of the boys actively participate in the drama club. If a student who participates in the drama club is chosen at random, what is the probability that they are a girl?
We have two cubic dice in a shaker. The first is a normal die with faces numbered from $1$ to $6$. The second is a biased die that has three faces with a $2$, two faces…
We have two cubic dice in a shaker. The first is a normal die with faces numbered from $1$ to $6$. The second is a biased die that has three faces with a $2$, two faces with a $3$ and one face with a $5$. A die is chosen at random (with probability $\frac{1}{2}$ for each) and rolled. If the result obtained is an even number, what is the probability that the biased die was rolled?
Back to the start: with 25 names in the hat, your chance of being picked first is $\frac{1}{25}$. If you weren't picked first, your chance of being picked second rises to $\frac{1}{24}$, because the draws are dependent. And the doctor? A positive test updates the probability with Bayes' Theorem, weighted by how common the condition is.