Section PC.4 – Independent Events; Conditional Probability
Independent Events
| Example 1 |
| Suppose we flipped a coin and rolled a die, and wanted to know the probability of the coin landing on heads and rolling a 6 on the die. |
| We could list all possible outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.
Notice there are 2 · 6 = 12 total outcomes. Out of these, only 1 is the desired outcome, so the probability is P(head and 6) = [latex]\frac{1}{12}[/latex]. |
The prior example was looking at two independent events. The events “landing on heads” and “rolling a 6” are independent because the outcome of the die roll does not affect the outcome of the coin flip.
| Independent Events |
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Events A and B are independent events if the probability of Event B occurring is the same whether or not Event A occurs. |
| Example 2 |
| Are these events independent?
a) A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head. b) You draw a card from a deck, then draw a second card without replacing the first. |
| a) The probability that a head comes up on the second toss is [latex]\frac{1}{2}[/latex] regardless of whether or not a head came up on the first toss, so these events are independent.
b) The probability of the first card being red is P(first card is red) = [latex]\frac{26}{52}[/latex] = [latex]\frac{1}{2}[/latex]. Since the first card is not returned to the deck, the probability of the second card being red is not the same as the probability that the first card is red. After drawing the first card, there are only 51 cards remaining in the deck. If the first card was red, then there are only 25 red cards remaining, and the probability that the second card would be red is P(second card is red) = [latex]\frac{25}{51}[/latex]. If the first card was not red, then the probability that the second card would be red is P(second card is red) = [latex]\frac{25}{51}[/latex]. These events are NOT independent. |
When two events are independent, the probability of both occurring is just the product of the probabilities of the individual events.
| P(A and B ) for independent events |
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If events A and B are independent, then the probability of both A and B occurring is P(A and B) = P(A) · P(B) where P(A and B ) is the probability of events A and B both occurring, P(A) is the probability of event A occurring, and P(B) is the probability of event B occurring. |
If you look back at the coin and die example from earlier, you can see how the number of outcomes of the first event multiplied by the number of outcomes in the second event multiplied to equal the total number of possible outcomes in the combined event.
| Example 3 |
| In Alex’s drawer she has 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If she randomly reaches in and pulls out a pair of socks and a tee shirt, what is the probability that both are white? |
| The probability of choosing a white pair of socks is The probability of both being white is [latex]\frac{6}{10}[/latex].
The probability of choosing a white tee shirt is [latex]\frac{3}{7}[/latex]. The probability of both being white is [latex]\frac{6}{10}[/latex] · [latex]\frac{3}{7}[/latex] = [latex]\frac{18}{70}[/latex] = [latex]\frac{9}{35}[/latex]. |
| You Try PC.4.A |
| A couple has two children. Find the probability that both children are girls. |
| Example 4 | ||||||||||||||||||||||||||||||||||||
| You roll two die. Find the probability that the first die lands on an even number, and the second die lands on 5. | ||||||||||||||||||||||||||||||||||||
| Since the outcome of the first die has no effect on the outcome of the second die, the events are independent.
P(first die is even) = [latex]\frac{3}{6}[/latex] = [latex]\frac{1}{2}[/latex] P(second die lands on 5) = [latex]\frac{1}{6}[/latex] P(first die is even AND second die lands on 5) = [latex]\frac{1}{2}[/latex] X [latex]\frac{1}{6}[/latex] = [latex]\frac{1}{12}[/latex] As a check, it may be helpful to consider all possible outcomes in the experiment of rolling 2 die:
Notice there are 3 outcomes (out of 36 possible outcomes) where the first die is even and the second die lands on 5. We could have found this probability directly as [latex]\frac{3}{36}[/latex] = [latex]\frac{1}{12}[/latex]. |
Conditional Probability
Often it is required to compute the probability of an event given that another event has occurred.
| Example 5 |
| What is the probability that two cards drawn at random from a deck of playing cards will both be aces? |
| It might seem that you could use the formula for the probability of two independent events and simply multiply [latex]\frac{4}{52}[/latex] · [latex]\frac{4}{52}[/latex] = [latex]\frac{1}{169}[/latex]. This would be incorrect, however, because the two events are not independent. Once the first card is drawn, there are only 51 cards remaining in the deck. If the first card drawn is an ace, then the probability that the second card is also an ace would be [latex]\frac{3}{51}[/latex] 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}[/latex] = [latex]\frac{1}{17}[/latex]. Thus, the probability of both cards being aces is [latex]\frac{4}{52}[/latex] · [latex]\frac{3}{51}[/latex] = [latex]\frac{12}{2652}[/latex] = [latex]\frac{1}{221}[/latex]. |
| Conditional Probability |
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The probability that 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” |
| Example 6 |
| Find the probability that a die rolled shows a 6, given that a flipped coin shows a head. |
| These are two independent events, so the probability of the die rolling a 6 is , regardless of the result of the coin flip. |
| Example 7 | ||||||||||||||||
The 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 cars. Find the probability that a randomly chosen person:
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| a) Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so
P(ticket | red car) = [latex]\frac{15}{150}[/latex] = [latex]\frac{1}{10}[/latex] = 0.1 b) Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so P(red car | ticket) = [latex]\frac{15}{60}[/latex] = [latex]\frac{1}{4}[/latex] = 0.25 |
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 an accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.
| P(A and B) for dependent events |
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If Events A and B are dependent (not independent), then P(A and B) = P(A) · P(B | A) |
| Example 8 |
| If you pull 2 cards out of a deck, what is the probability that both are spades? |
| 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}[/latex] · [latex]\frac{12}{51}[/latex] ≈ 0.0588 |
| You Try PC.4.B |
| If you draw 2 cards from a standard deck, what is the probability that both cards are red kings? |
| Example 9 | ||||||||||||||||
| A home pregnancy test was given to women, then pregnancy was verified through blood tests. The following table shows the home pregnancy test results. Find
a) P(not pregnant | positive test result) b) P(positive test result | not pregnant)
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| a) Since we know the test result was positive, we’re limited to the 75 women in the first column, of which 5 were not pregnant.
P(not pregnant | positive test result) = [latex]\frac{5}{75}[/latex] ≈ 0.067. b) Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test. P(positive test result | not pregnant) = [latex]\frac{5}{19}[/latex] ≈ 0.263. The second result is what is usually called a false positive: A positive result when the woman is not actually pregnant. |
| Example 10 | ||||||||||||||||
| A certain virus infects 100 in every 6,000 people. A test used to detect the virus in a person is positive 94% of the time if the person has the virus and 8% of the time if the person does not have the virus. Let A be the event “the person is infected” and B be the event “the person tests positive”.
a. Find the probability that a person has the virus given that they have tested positive, i.e. find P(A | B). b. Find the probability that a person does not have the virus given that they test negative, i.e. find P(not A | not B). c. Find the probability that a person has the virus given that they have tested positive, i.e. find P(A | B).
P( Person has the virus | Tested Positive) = [latex]\frac{94}{5640}[/latex] = 0.0166666667 The probability that a person has the virus given that they have tested positive is 0.0167 d. a. Find the probability that a person does not have the virus given that they test negative, i.e. find P(not A | not B). P(Person does not have the virus | Tested Negative) = [latex]\frac{354}{360}[/latex] = 0.9833 |
Section PC.4 Answers to You Try Problems
PC.4.A
[latex]\frac{1}{4}[/latex]
PC.4.B
0.00075
