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Basic Ideas in Probability Study Guide (page 3)

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Updated on Oct 5, 2011

Example

A candy store manager has kept a record of the purchases his customers make. He has determined the probability that a randomly selected customer will purchase gum is 0.4; the probability that a randomly selected customer will buy chocolate candy is 0.7; and the probability the customer purchases both is 0.2.

  1. What is the probability that a randomly selected customer buys gum or chocolate candy?
  2. What is the probability that a randomly selected customer will not buy gum?
  3. What is the probability that a randomly selected customer will buy neither gum nor chocolate candy?

Solution

Let G = the event a customer purchases gum and C = the event a customer purchases chocolate candy. We are told P(G) = 0.4, P(C) = 0.7, and P(G and C) = 0.2. Hint: It always helps to write down what is given using proper notation.

  1. We want to find the probability of G or C occurring, so we will use axiom 6: P(G or C) = P(G) + P(C) – P(G and C) = 0.4 + 0.7 – 0.2 = 0.9
  2. Not buying gum is the complement of buying gum. Thus: P(not G) = 1 – P(G) = 1 – 0.4 = 0.6
  3. The event of not buying either gum or chocolate candy is the complement of buying one or the other or both. That is: P(not G and not C) = 1 – P(G or C) = 1 – 0.9 = 0.1

Conditional Probability and Independence

Sometimes, knowing that one event has occurred changes our knowledge about the likelihood that another event has occurred. For example, suppose a family with two children is moving into the house across the street.What is the probability that the family has two girls? Using B to denote a boy and G to represent a girl, the sample space for this experiment is S = {BB, BG, GB, GG}. Note that both BG and GB are outcomes in the sample space because BG represents the outcome that the older child is a boy, GB implies the older child is a girl. For all of the outcomes in the sample space to be equally likely, both must be listed. The probability of two girls is because only one of the four outcomes in the sample space has two girls.

Now suppose that we learn that the family has at least one girl.We know that the family does not have two boys, so the sample space is reduced to S2 = {BG, GB, GG}. Each of the outcomes remains equally likely, so the probability of two girls knowing that the family has at least one girl is .

If the number of possible outcomes is small, as was the case in this example, it is not difficult to determine both the sample space for the experiment and the reduced sample space given some information. However, as the size of the sample space increases, this approach quickly becomes impractical.We need a general approach to determining conditional probabilities. In general, for any event F such that P(F) > 0, the conditional probability of the event E given that the event F has occurred is:

For our family with two children, E would be the event of two girls and F would be the event of at least one girl. If a family must have two girls and at least one girl, then they have two girls (E cannot occur if the family has only one girl). Thus, the probability of E and F both occurring is , the probability of two girls. The probability F occurs is because three of the four outcomes in the sample space have at least one girl. Therefore, , which agrees with our earlier result based on the reduced sample space.

Sometimes, knowledge that one event has occurred does not change the probability that another event has occurred; that is, P(E|F) = P(E). Now, let us rewrite as P(E and F) = P(E|F)P(F). If E and F are independent, P(E|F) = P(E) and thus P(E and F) = P(E)P(F). When P(E and F) = P(E)P(F) or, equivalently, P(E|F) = P(E), the events E and F are said to be independent events because knowledge of whether or not the event F has occurred has no effect on the probability that E has occurred. Further, given that E and F are independent events, P(E and F) = P(E)P(F) and P(E|F) = P(E).

Conditional Probability

In general, the conditional probability of the event E given that the event F has occurred is:

Example

In a housing development, the steering committee wanted to examine the need for pet and child play areas. The head of each household in the division was asked how many children lived in the household and whether or not they had a dog. The results are shown in Table 9.1.

Table 9.1 Number of children and dogs per family

  1. What is the probability that a randomly selected household in this subdivision has a dog?
  2. What is the probability that a randomly selected household in this subdivision has children?
  3. What is the probability that a randomly selected household in this subdivision has a dog and one child?
  4. What is the probability that a randomly selected household in this subdivision has a dog given that one child is in it?
  5. Is the event of having a dog independent of having one child for households in this neighborhood?

Solution

Let D = event the household has a dog. Let A = event the household has no children, B = event the household has one child, C = event the household has at least two children, and E = event the household as at least one child.

  1. Since 132 of the 156 households have dogs, .
  2. A household has children if it has one or more children. Thus, is the probability that the randomly selected household has at least one child.
  3. Forty-six households had both a dog and one child, so .
  4. Of the 94 households with one child, 46 had a dog. Therefore, .
  5. Having a dog is not independent of having one child in this housing division because .

Theorem of Total Probability

Suppose that the sample space is partitioned into n mutually exclusive and exhaustive events, Ei, i = 1, 2, . . . , n, as in Figure 9.2. By mutually exclusive,we mean that the events have no outcomes in common. By exhaustive, we mean that, when combining the outcomes from all events,we have all points in the sample space. Now suppose that we are interested in determining the probability of the event A. It may be difficult to find the probability of A directly, but we can sometimes use the partitioning of the sample space to make this easier. The Theorem of Total Probability states that, if Ei, i = 1, 2, . . . , n are disjoint events, with P(E1) + P(E2) + . . . + P(En) = 1, then for any event F:

P(F) = P(F and E1) + P(F and E2) + ... + P(F and En) = P(F|E1)P(E1) + P(F|E2)P(E2) + ... + P(F|En)P(En)

Figure 9.2

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