The conditional probability of A given B must be at least as large as the probability of A
Learning Objectives Suppose a fair die has been rolled and you are asked to give the probability that it was a five. There are six equally likely outcomes, so your answer is \(1/6\). But suppose that before you give your answer you are given the extra information that the number rolled was odd. Since there are only three odd numbers that are possible, one of which is five, you would certai: nly revise your estimate of the likelihood that a five was rolled from \(1/6\) to \(1/3\). In general, the revised probability that an event A has occurred, taking into account the additional information that another event \(B\) has definitely occurred on this trial of the experiment, is called the conditional probability of \(A\) given \(B\) and is denoted by \(P(A\mid B)\). The reasoning employed in this example can be generalized to yield the computational formula in the following definition. Definition: conditional probability The conditional probability of \(A\) given \(B\), denoted \(P(A\mid B)\), is the probability that event \(A\) has occurred in a trial of a random experiment for which it is known that event \(B\) has definitely occurred. It may be computed by means of the following formula: \[P(A\mid B)=\dfrac{P(A\cap B)}{P(B)} \label{CondProb}\] Example \(\PageIndex{1}\): Rolling a Die A fair (unbiased) die is rolled.
Solution: The sample space for this experiment is the set \(S={1,2,3,4,5,6}\) consisting of six equally likely outcomes. Let \(F\) denote the event “a five is rolled” and let \(O\) denote the event “an odd number is rolled,” so that \[F={5}\; \; \text{and}\; \; O={1,3,5} \nonumber\]
Just as we did not need the computational formula in this example, we do not need it when the information is presented in a two-way classification table, as in the next example. Example \(\PageIndex{2}\): Marriage and Gender In a sample of \(902\) individuals under \(40\) who were or had previously been married, each person was classified according to gender and age at first marriage. The results are summarized in the following two-way classification table, where the meaning of the labels is:
The numbers in the first row mean that \(43\) people in the sample were men who were first married in their teens, \(293\) were men who were first married in their twenties, \(114\) men who were first married in their thirties, and a total of \(450\) people in the sample were men. Similarly for the numbers in the second row. The numbers in the last row mean that, irrespective of gender, \(125\) people in the sample were married in their teens, \(592\) in their twenties, \(185\) in their thirties, and that there were \(902\) people in the sample in all. Suppose that the proportions in the sample accurately reflect those in the population of all individuals in the population who are under \(40\) and who are or have previously been married. Suppose such a person is selected at random.
Solution: It is natural to let \(E\) also denote the event that the person selected was a teenager at first marriage and to let \(M\) denote the event that the person selected is male.
The proportion of males in the sample who were in their teens at their first marriage is \(43/450\). This is the relative frequency of such people in the population of males, hence \(P(E/M)=43/450\approx 0.096\) or about \(10\%\). In the next example, the computational formula in the definition must be used. Example \(\PageIndex{3}\): Body Weigth and hypertension Suppose that in an adult population the proportion of people who are both overweight and suffer hypertension is \(0.09\); the proportion of people who are not overweight but suffer hypertension is \(0.11\); the proportion of people who are overweight but do not suffer hypertension is \(0.02\); and the proportion of people who are neither overweight nor suffer hypertension is \(0.78\). An adult is randomly selected from this population.
Solution: Let \(H\) denote the event “the person selected suffers hypertension.” Let \(O\) denote the event “the person selected is overweight.” The probability information given in the problem may be organized into the following contingency table:
Independent EventsAlthough typically we expect the conditional probability \(P(A\mid B)\) to be different from the probability \(P(A)\) of \(A\), it does not have to be different from \(P(A)\). When \(P(A\mid B)=P(A)\), the occurrence of \(B\) has no effect on the likelihood of \(A\). Whether or not the event \(A\) has occurred is independent of the event \(B\). Using algebra it can be shown that the equality \(P(A\mid B)=P(A)\) holds if and only if the equality \(P(A\cap B)=P(A)\cdot P(B)\) holds, which in turn is true if and only if \(P(B\mid A)=P(B)\). This is the basis for the following definition. Definition: Independent and Dependent Events Events \(A\) and \(B\) are independent (i.e., events whose probability of occurring together is the product of their individual probabilities). if \[P(A\cap B)=P(A)\cdot P(B)\] If \(A\) and \(B\) are not independent then they are dependent. The formula in the definition has two practical but exactly opposite uses:
Example \(\PageIndex{4}\): Rolling a Die again A single fair die is rolled. Let \(A=\{3\}\) and \(B=\{1,3,5\}\). Are \(A\) and \(B\) independent? Solution: In this example we can compute all three probabilities \(P(A)=1/6\), \(P(B)=1/2\), and \(P(A\cap B)=P(\{3\})=1/6\). Since the product \(P(A)\cdot P(B)=(1/6)(1/2)=1/12\) is not the same number as \(P(A\cap B)=1/6\), the events \(A\) and \(B\) are not independent. Example \(\PageIndex{5}\) The two-way classification of married or previously married adults under \(40\) according to gender and age at first marriage produced the table
Determine whether or not the events \(F\): “female” and \(E\): “was a teenager at first marriage” are independent. Solution: The table shows that in the sample of \(902\) such adults, \(452\) were female, \(125\) were teenagers at their first marriage, and \(82\) were females who were teenagers at their first marriage, so that \[ \begin{align*} P(F) &=\dfrac{452}{902},\\[4pt] P(E) &=\dfrac{125}{902} \\[4pt] P(F\cap E) &=\dfrac{82}{902} \end{align*}\] Since \[ \begin{align*} P(F)\cdot P(E) &=\dfrac{452}{902}\cdot \dfrac{125}{902} \\[4pt] &=0.069 \end{align*}\] is not the same as \[P(F\cap E)=\dfrac{82}{902}=0.091 \nonumber\] we conclude that the two events are not independent. Example \(\PageIndex{6}\) Many diagnostic tests for detecting diseases do not test for the disease directly but for a chemical or biological product of the disease, hence are not perfectly reliable. The sensitivity of a test is the probability that the test will be positive when administered to a person who has the disease. The higher the sensitivity, the greater the detection rate and the lower the false negative rate. Suppose the sensitivity of a diagnostic procedure to test whether a person has a particular disease is \(92\%\). A person who actually has the disease is tested for it using this procedure by two independent laboratories.
Solution:
Example \(\PageIndex{7}\): specificity of a diagnostic test The specificity of a diagnostic test for a disease is the probability that the test will be negative when administered to a person who does not have the disease. The higher the specificity, the lower the false positive rate. Suppose the specificity of a diagnostic procedure to test whether a person has a particular disease is \(89\%\).
Solution:
The concept of independence applies to any number of events. For example, three events \(A,\; B,\; \text{and}\; C\) are independent if \(P(A\cap B\cap C)=P(A)\cdot P(B)\cdot P(C)\). Note carefully that, as is the case with just two events, this is not a formula that is always valid, but holds precisely when the events in question are independent. Example \(\PageIndex{8}\): redundancy The reliability of a system can be enhanced by redundancy, which means building two or more independent devices to do the same job, such as two independent braking systems in an automobile. Suppose a particular species of trained dogs has a \(90\%\) chance of detecting contraband in airline luggage. If the luggage is checked three times by three different dogs independently of one another, what is the probability that contraband will be detected? Solution: Let \(D_1\) denote the event that the contraband is detected by the first dog, \(D_2\) the event that it is detected by the second dog, and \(D_3\) the event that it is detected by the third. Since each dog has a \(90\%\) of detecting the contraband, by the Probability Rule for Complements it has a \(10\%\) chance of failing. In symbols, \[P(D_{1}^{c})=0.10,\; \; P(D_{2}^{c})=0.10,\; \; P(D_{3}^{c})=0.10\] Let \(D\) denote the event that the contraband is detected. We seek \(P(D)\). It is easier to find \(P(D^c)\), because although there are several ways for the contraband to be detected, there is only one way for it to go undetected: all three dogs must fail. Thus \(D^c=D_{1}^{c}\cap D_{2}^{c}\cap D_{3}^{c}\) and \[P(D)=1-P(D^c)=1-P(D_{1}^{c}\cap D_{2}^{c}\cap D_{3}^{c})\]But the events \(D_1\), \(D_2\), and \(D_3\) are independent, which implies that their complements are independent, so \[P(D_{1}^{c}\cap D_{2}^{c}\cap D_{3}^{c})=P(D_{1}^{c})\cdot P(D_{2}^{c})\cdot P(D_{3}^{c})=0.10\times 0.10\times 0.10=0.001\] Using this number in the previous display we obtain \[P(D)=1-0.001=0.999\] That is, although any one dog has only a \(90\%\) chance of detecting the contraband, three dogs working independently have a \(99.9\%\) chance of detecting it. Probabilities on Tree DiagramsSome probability problems are made much simpler when approached using a tree diagram. The next example illustrates how to place probabilities on a tree diagram and use it to solve a problem. Example \(\PageIndex{9}\): A jar of Marbles A jar contains \(10\) marbles, \(7\) black and \(3\) white. Two marbles are drawn without replacement, which means that the first one is not put back before the second one is drawn.
Solution: A tree diagram for the situation of drawing one marble after the other without replacement is shown in Figure \(\PageIndex{1}\). The circle and rectangle will be explained later, and should be ignored for now. Figure \(\PageIndex{1}\): Tree Diagram for Drawing Two MarblesThe numbers on the two leftmost branches are the probabilities of getting either a black marble, \(7\) out of \(10\), or a white marble, \(3\) out of \(10\), on the first draw. The number on each remaining branch is the probability of the event corresponding to the node on the right end of the branch occurring, given that the event corresponding to the node on the left end of the branch has occurred. Thus for the top branch, connecting the two Bs, it is \(P(B_2\mid B_1)\), where \(B_1\) denotes the event “the first marble drawn is black” and \(B_2\) denotes the event “the second marble drawn is black.” Since after drawing a black marble out there are \(9\) marbles left, of which \(6\) are black, this probability is \(6/9\). The number to the right of each final node is computed as shown, using the principle that if the formula in the Conditional Rule for Probability is multiplied by \(P(B)\), then the result is \[P(B\cap A)=P(B)\cdot P(A\mid B)\]
Of course, this answer could have been found more easily using the Probability Law for Complements, simply subtracting the probability of the complementary event, “two white marbles are drawn,” from 1 to obtain \(1-0.07=0.93\). As this example shows, finding the probability for each branch is fairly straightforward, since we compute it knowing everything that has happened in the sequence of steps so far. Two principles that are true in general emerge from this example: Probabilities on Tree Diagrams
Key Takeaway
What is the conditional probability of a given a?Conditional probability refers to the chances that some outcome occurs given that another event has also occurred. It is often stated as the probability of B given A and is written as P(B|A), where the probability of B depends on that of A happening.
What is the conditional probability of event A given event B has occurred?If A and B are two events in a sample space S, then the conditional probability of A given B is defined as P(A|B)=P(A∩B)P(B), when P(B)>0.
How do you find the probability of B given a?This probability is written P(B|A), notation for the probability of B given A. In the case where events A and B are independent (where event A has no effect on the probability of event B), the conditional probability of event B given event A is simply the probability of event B, that is P(B). P(A and B) = P(A)P(B|A).
How do you do probability when at least?To calculate the probability of an event occurring at least once, it will be the complement of the event never occurring. This means that the probability of the event never occurring and the probability of the event occurring at least once will equal one, or a 100% chance.
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