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5.2: Probability: Living with odds

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    4918
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    Probability is a subtle concept: There are several different things we mean by probable.

    Randomness is also subtle. Ten heads in a row? Did we cheat? We probably did, but don’t know yet as of when we typed this. In order to get ten heads in a row, we expect to have to cheat, but we don't know for sure that we will have to.

    Our knowledge of things to come is also imperfect. What can we say in the face of imperfect knowledge? How can we reason knowing our knowledge is imperfect?

    Definitions:

    Outcome: An outcome (simple outcome) is the most basic possible result of observations or experiments.

    Sample Space: Set of all possible outcomes.

    Events: An event is one or more outcomes that share a property of interest

    Example \(\PageIndex{1}\):

    In a coin toss, the possible outcomes are T, H.

    In tossing two coins the outcomes are TT, TH, HT, HH.

    In rolling a standard die the outcomes are 1, 2, 3, 4, 5 or 6 dots.

    In an experiment on colours of bred pea plants, the outcomes could be a red, white or pink plant.

    Types of Probability:

    Experimental: We observe over a length of time or perform an experiment many times and calculate the relative frequency of the event. The relative frequency is the number of times the desired outcome occurs per the number of times an experiment is performed or observations made – it’s a percentage!

    Theoretical: Based on a model where all outcomes are equally likely.

    Subjective: Estimate based on intuition or experience (ideally to be made by an expert in the field who also has a sound grasp of probability).

    Fundamentals of Probability

    Example \(\PageIndex{2}\):

    • When tossing two coins we might be interested in the event of two heads, or the event of at least one head.
    • When rolling a standard die we might be interested in the event of rolling a one or the event of rolling an even number or the event of rolling a number less than 4.
    • When breeding 6 pea plants we might be interested in the event all plants are the same colour.

    Example \(\PageIndex{3}\):

    Experiment: Toss two coins

    Outcomes: TT, TH, HT, HH

    Event: getting one head consists of TH or HT

    Simple Calculations

    Assumptions: a fair coin, a fair dice, well-shuffled cards

    Method: Count the total number of possible outcomes \(n\) and count the number of outcomes in the event \(A\).

    \(P(A) = \displaystyle \frac{|A|}{n}\).

    In general, the probability of an event, \(A\), occurring is \(P(A) = \displaystyle \frac{|A|}{|S|}\), where \(|S| \) is the total number of outcomes in the sample space \(S\).

    Formal Properties of Probability

    Rules:

    Let A, B, C, … be events. P(A) denotes the probability event A occurs.

    RULE #1: \(0 \leq P(A) \leq 1\).for every event \(A\). The probability of any event is a number between 0 and 1.

    Rule # 2: P(E) = 0 if and only if the event is impossible.

    Rule# 3: P(E) = 1 if and only if the event is a certainty.

    Rule #4: \( P( E^c) = 1 – P(E)\). (The probability of an event NOT occurring is 1 – probability the event occurs).

    Thinking Out Loud:

    Probability can also be expressed as a percentage. What is the range?

    Example \(\PageIndex{4}\):

    What is the probability of rolling a four on two six-sided dice?

    There are 36 different ways of rolling two six-sided dice. (How do we know?)

    There are 3 ways of rolling four (1+3, 2+2, 3+1)

    So, the probability is 3/36 = 1/12

    Thinking Out Loud:

    What is the probability of flipping 10 coins and getting all the same?

    What is the probability of flipping 10 coins and getting all heads?

    Example \(\PageIndex{5}\):

    From a well-shuffled deck of \(52\) cards, three cards are drawn at random. Find the probability that exactly one King will be drawn.

    There are \(52 \choose 3\) ways to select \(3\) cards out of \(52\). There are \( 4\choose 1\) ways to select one King out of four Kings. Further, there are \(48 \choose 2\) ways to select the remaining two non-King cards out of \(48\) non-King cards. Therefore, the probability that exactly one King will be drawn is

    \[ \frac{\binom{4 } {1} \cdot \binom{48} {2}}{ 52 \choose 3}\].

    Example \(\PageIndex{6}\):

    From a well-shuffled deck of \(52\) cards, three cards are drawn at random. Find the probability that at least one King will be drawn.

    The probability is the sum of the probability that exactly one, two, and three King will be drawn. Hence ,\[ \frac{\binom{4 } {1} \cdot \binom{48} {2}}{ 52 \choose 3} + \frac{\binom{4 } {2} \cdot \binom{48} {1}}{ 52 \choose 3} + \frac{\binom{4 } {3} \cdot \binom{48} {0}}{ 52 \choose 3}\].

    Probability distribution alt

    The probability distribution is a display of the probability for every possible event. For example, see the probability distribution of rolling two six-sided dice.

    250px-Dice_Distribution_(bar).svg.png

    Combining Probabilities

    Definition: Independent

    Two events are independent if the outcome of one event does not affect the probability of the second occurring otherwise we say the two events are dependent

    Example \(\PageIndex{7}\):

    Tossing a coin twice, the event of the first toss being a head and the event of the second toss being a head are independent.

    Tossing a coin twice, the event of the first toss being a head and the event of at least one head occurring are NOT independent.

    Rules Continued:

    RULE #5: If A and B are independent events then P(A and B) = P(A)P(B).

    This rule generalizes to 3 or more independent events.:

    Example \(\PageIndex{8}\):

    Suppose you toss three coins. What is the probability of getting three tails? P(3 tails)=P(1tail) P(1tail) P(1tail)= 1/8.

    Rules Continued:

    RULE #6: If A and B are dependent events then P(A and B) = P(A)P(B given A)

    {P(B given A) is the conditional probability of even B occurring when event A is assumed to have already occurred).}

    Example \(\PageIndex{9}\):

    Two members are selected from a pool of 17 male students and 23 female students. Find the probability that the first student selected is a male and the second is also male.

    \(P(A) =\) The first student selected is male.

    \(P(B) =\) The second student selected is male

    \(P(A) = \displaystyle \frac{17}{40}\)

    \(P(B \, given \, A) = \displaystyle \frac{16}{39}\)

    \(PA \, and \, B) = \left( \displaystyle \frac{17}{40} \right) \left( \displaystyle \frac{16}{39} \right) \)

    Definition: Mutually Exclusive

    Two events are mutually exclusive (non-overlapping) if the two events can’t occur at the same time. Two events are overlapping if they can occur at the same time.

    Example \(\PageIndex{10}\):

    Tossing a coin twice, the event of the first toss being a head and the event of the second toss being a head are overlapping. The outcome of HH would satisfy both events. Tossing a coin twice, the event of the first toss being a head and the event of no heads occurring are non-overlapping, since we can’t have both occurring. When tossing a die, the event of tossing a number greater than 4 and the event of tossing an even number are overlapping events. An outcome of 6 would satisfy both.

    Rules Continued:

    Rule #7: If A and B are non-overlapping events then

    P(A or B) = P(A)+ P(B)

    This generalizes to 3 or more events.

    Example \(\PageIndex{11}\):

    Find the probability of rolling either a 1 or a 2 on a single die.

    \(P(1) = \displaystyle \frac{1}{6}\)

    \(P(2) = \displaystyle \frac{1}{6}\)

    \(P(1 \, or \, 2) = \displaystyle \frac{1}{6} + \displaystyle \frac{1}{6}\)

    \(P(1 \, or \, 2) = \displaystyle \frac{2}{6} = \displaystyle \frac{1}{3}\)

    Rules Continued:

    Rule #8: If A and B are overlapping events then

    P(A or B) = P(A)+ P(B) - P(A and B).

    Example \(\PageIndex{12}\):

    Find the probability of drawing a queen or a heart from a standard deck.

    \(P(Q) = \displaystyle \frac{4}{52}\)

    \(P(H) = \displaystyle \frac{13}{52}\)

    \(P(Q \, and \, H) = \displaystyle \frac{1}{52}\)

    \(P(Q \, or \, H) = \displaystyle \frac{4}{52} + \displaystyle \frac{13}{52} - \displaystyle \frac{1}{52}\)

    \(P(Q \, or \, H) = \displaystyle \frac{16}{52} = \displaystyle \frac{4}{13}\)

    Rules Continued:

    Rule #9: Suppose the probability of an event in one trial is P(A). If all trials are independent, the probability that event occurs at least once in n trials is

    \(1 - P \left(no \, event \, A \, in \, n \, trials \right) = 1 - P \left( not \, A \, in \, one \, trial \right) ^n \)

    Example \(\PageIndex{13}\):

    Find the probability of getting at least one 6 in five rolls of a single die.

    More on conditional probability

    We have seen that P(A and B) = P(A)P(B given A)

    I.e. P(B given A) = P(A and B)/P(A).

    P(B given A) = the (conditional) probability that event B occurs given that event A is known to have occurred.

    Example \(\PageIndex{14}\):

    if a coin was tossed twice, P(HH) =\(\frac{1}{2}\).

    P(HH given both coins showed same value) = \(\frac{1}{4}\). ( the condition tells us it was either HH or TT).

    Rules Continued:

    Rule #10: The odds for event \(A\) occurring are:

    Odds for event \(A = P(A):P(not A)\). This is usually converted to a pair of integers.

    Example \(\PageIndex{15}\):

    If you toss a coin twice, what are the odds you get at least one head appearing?

    1:1 (50:50)

    If the odds are 1:3 that a patient will survive heart surgery is that good or bad? What is it as a probability?

    ¼ = 0.25 = 25% not great

    Example \(\PageIndex{16}\):

    A fair coin is tossed 10 times. Which is more likely:

    1. 10 heads in a row
    2. THHTHTHTTH (giving 5 heads and 5 tails in total).

    Both events are equally as likely- in each case, I explicitly predicted each toss of the coin! We confuse the fact that it is more likely to get 5 heads than 10 heads with a very specific outcome.

    Coincidences

    Partly because we don’t know probabilities as well as we should and partly because we over publicize uncommon events and attach significance to things after they happen we don’t appreciate coincidences properly.

    When is your birthday: month and day?

    It can be shown that in a room with 23 people there is a 50% chance two will share a birthday. In a room with 41 or more people, the probability is over 90%. (Note: if we specified in advance which day the birthday was on it would be far less likely to find two people with a birthday on that day).

    Monty Hall Problem

    Monty Hall runs a game show. Contestants pick one of three closed doors and win the prize behind the door. Behind one of the doors is a car, behind the other two doors, are pigs. After the contestant picks a door, Monty Hall opens one of the other doors to reveal one of the pigs. He asks the contestant if they would like to change their mind and take the other unopened door. Should they bother switching? Why?


    This page titled 5.2: Probability: Living with odds is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Pamini Thangarajah.

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