Exercise: What is the probability that two cards drawn from the deck are the same suit? Different suits? Answer these questions for both the case of choosing the cards with and without replacement. Show Competencies: Calculate the probability that four cards dealt from a deck without replacement are of different suits, both by conditional probability and by counting arguments. Reflection: Which problems are more easily done as conditional probability and which problems are more easily done by counting arguments. Challenge: Calculate the probability that three cards dealt from a deck without replacement are of different suits, both by conditional probability and by counting arguments. In the previous section we computed the probabilities of events that were independent of each other. We saw that getting a certain outcome from rolling a die had no influence on the outcome from flipping a coin, even though we were computing a probability based on doing them at the same time. In this section, we will consider events that are dependent on each other, called conditional probabilities. Conditional ProbabilityThe probability the event B occurs, given that event A has happened, is represented as P(B | A) This is read as “the probability of B given A” For example, if you draw a card from a deck, then the sample space for the next card drawn has changed, because you are now working with a deck of 51 cards. In the following example we will show you how the computations for events like this are different from the computations we did in the last section. exampleWhat is the probability that two cards drawn at random from a deck of playing cards will both be aces? Show Solution It might seem that you could use the formula for the probability of two independent events and simply multiply [latex]\frac{4}{52}\cdot\frac{4}{52}=\frac{1}{169}[/latex]. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck. Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case the “condition” is that the first card is an ace. Symbolically, we write this as: P(ace on second draw | an ace on the first draw). The vertical bar “|” is read as “given,” so the above expression is short for “The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw.” What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is [latex]\frac{3}{51}=\frac{1}{17}[/latex]. Thus, the probability of both cards being aces is [latex]\frac{4}{52}\cdot\frac{3}{51}=\frac{12}{2652}=\frac{1}{221}[/latex]. Conditional Probability FormulaIf Events A and B are not independent, then P(A and B) = P(A) · P(B | A) exampleIf you pull 2 cards out of a deck, what is the probability that both are spades? Show Solution The probability that the first card is a spade is [latex]\frac{13}{52}[/latex]. The probability that the second card is a spade, given the first was a spade, is [latex]\frac{12}{51}[/latex], since there is one less spade in the deck, and one less total cards. The probability that both cards are spades is [latex]\frac{13}{52}\cdot\frac{12}{51}=\frac{156}{2652}\approx0.0588[/latex] Try ItExampleThe table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:
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Notice from the last example that P(B | A) is not equal to P(A | B). These kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look at the conditional probability of you having accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood. View more about conditional probability in the following video.
ExampleIf you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card? Show Solution You can satisfy this condition by having Case A or Case B, as follows: Case A) you can get the Ace of Diamonds first and then a black card or Case B) you can get a black card first and then the Ace of Diamonds. Let’s calculate the probability of Case A. The probability that the first card is the Ace of Diamonds is [latex]\frac{1}{52}[/latex]. The probability that the second card is black given that the first card is the Ace of Diamonds is [latex]\frac{26}{51}[/latex] because 26 of the remaining 51 cards are black. The probability is therefore [latex]\frac{1}{52}\cdot\frac{26}{51}=\frac{1}{102}[/latex]. Now for Case B: the probability that the first card is black is [latex]\frac{26}{52}=\frac{1}{2}[/latex]. The probability that the second card is the Ace of Diamonds given that the first card is black is [latex]\frac{1}{51}[/latex]. The probability of Case B is therefore [latex]\frac{1}{2}\cdot\frac{1}{51}=\frac{1}{102}[/latex], the same as the probability of Case 1. Recall that the probability of A or B is P(A) + P(B) – P(A and B). In this problem, P(A and B) = 0 since the first card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is [latex]\frac{1}{101}+\frac{1}{101}=\frac{2}{101}[/latex]. The probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck is [latex]\frac{2}{101}[/latex]. These two playing card scenarios are discussed further in the following video. Try ItExampleA home pregnancy test was given to women, then pregnancy was verified through blood tests. The following table shows the home pregnancy test results. Find
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The second result is what is usually called a false positive: A positive result when the woman is not actually pregnant. What is the probability of getting a non ace?Hence the probability of getting a non-ace card is 12/13.
What is the probability of a card being an ace given that it is a nonThe probability of it being an Ace given it is a Non-face card uses the conditional formula: Note, that in a standard deck of 52 cards, there are 12 face cards, so 40 non-face cards. Of those non-face cards, there are only 4 Aces.
What is the probability that the card is an ace?The probability of getting an ace is 1/52.
What is the probability of picking a nonTherefore, there are 12 Face cards in total (6 red and 6 black), 4 Aces. 4 Kings (2 red and 2 black), 4 Queens (2 red and 2 black). Therefore, the probability of finding a non-face card in a well shuffled deck of 52 playing cards is 1013 .
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