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Chapter 4: Problem 68

Given that \(A, B\), and \(C\) are three independent events, find their joint probability for the following. a. \(P(A)=.81, \quad P(B)=.49\), and \(P(C)=.36\) b. \(P(A)=.02, \quad P(B)=.03\), and \(P(C)=.05\)

Short Answer

Expert verified
The joint probabilities for the given individual probabilities of events A, B, and C are as follows: for the first case is 0.81 * 0.49 * 0.36 and for the second case is 0.02 * 0.03 * 0.05

Step by step solution

01

Understand the concept of independent events

Two or more events are said to be independent if the occurrence of any one event doesn't affect the occurrence of the other event(s). This means that for independent events A, B and C, their joint probability is calculated as the product of their individual probabilities, i.e, P(A and B and C) = P(A) * P(B) * P(C).
02

Calculate the joint probability for the first set of probabilities

Use the given probabilities of individual events A, B, and C for the first case as 0.81, 0.49, and 0.36 respectively. Substitute these values in the formula to get the joint probability. So, \(P(A and B and C) = P(A) * P(B) * P(C) = 0.81 * 0.49 * 0.36.\)
03

Calculate the joint probability for the second set of probabilities

For the second case, the given probabilities are 0.02, 0.03, and 0.05 respectively for events A, B, and C. Again, substitute these in the formula. So, \(P(A and B and C) = P(A) * P(B) * P(C) = 0.02 * 0.03 * 0.05.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Events
To grasp the idea of independent events, think of two or more occurrences where the outcome of one event does not impact the others. For instance, imagine tossing a coin and rolling a dice at the same time. The result of the coin toss does not affect the dice roll outcome; thus, these events are independent.
This notion is foundational in probability theory. When events are independent, their joint probability鈥攐r the chance that all events happen together鈥攃an be simply calculated by multiplying their individual probabilities.
Independent events allow for simplifying complex probability calculations, which is why understanding this concept is important.
Probability Calculation
Calculating probability involves assigning a number between 0 and 1 to describe the likelihood of an event happening. A probability of 0 means the event cannot happen, while a probability of 1 implies certainty.
To calculate the probability of independent events, we consider each event separately at first. Each event is assigned a probability based on the likelihood of its occurrence. For example, if a coin is tossed, the probability of landing on heads is 0.5.
  • The total probability must always add up to 1 when considering all possible outcomes.
  • Make sure each probability is between 0 and 1.
These considerations help ensure the accuracy of calculations involving independent events.
Multiplication Rule for Probability
The multiplication rule is a straightforward yet effective method for finding joint probabilities of independent events. It states that if events are independent, the probability of all events occurring together can be found by multiplying their individual probabilities.
This rule works because independence implies no influence between events.
  • Example: If event A has a probability of 0.81, event B has a probability of 0.49, and event C has a probability of 0.36, then the joint probability is calculated as: \[ P(A \text{ and } B \text{ and } C) = P(A) \times P(B) \times P(C) = 0.81 \times 0.49 \times 0.36 \]
  • Remember, this formula is only applicable to independent events.
Using the multiplication rule simplifies finding joint probabilities and illustrates why knowing both the probabilities and the independence of events is crucial.

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Most popular questions from this chapter

Recent uncertain economic conditions have forced many people to change their spending habits. In a recent telephone poll of 1000 adults, 629 stated that they were cutting back on their daily spending. Suppose that 322 of the 629 people who stated that they were cutting back on their daily spending said that they were cutting back "somewhat" and 97 stated that they were cutting back "somewhat" and "delaying the purchase of a new car by at least 6 months". If one of the 629 people who are cutting back on their spending is selected at random, what is the probability that he/she is delaying the purchase of a new car by at least 6 months given that he/she is cutting back on spending "somewhat?"

Refer to Exercise 4.52, which contains information on a July 21, 2009 www.HuffingtonPost.com survey that asked people to choose their favorite junk food from a list of choices. The following table contains results classified by gender. (Note: There are 4801 females and 3201 males.) $$ \begin{array}{lcc} \hline \text { Favorite Junk Food } & \text { Female } & \text { Male } \\ \hline \text { Chocolate } & 1518 & 531 \\ \text { Sugary candy } & 218 & 127 \\ \text { Ice cream } & 685 & 586 \\ \text { Fast food } & 312 & 463 \\ \text { Cookies } & 431 & 219 \\ \text { Chips } & 458 & 649 \\ \text { Cake } & 387 & 103 \\ \text { Pizza } & 792 & 523 \\ \hline \end{array} $$ a. Suppose that one person is selected at random from this sample of 8002 respondents. Find the following probabilities. i. Probability of the intersection of events female and ice cream. ii. Probability of the intersection of events male and pizza. b. Mention at least four other joint probabilities you can calculate for this table and then find their probabilities. You may draw a tree diagram to find these probabilities.

Given that \(A\) and \(B\) are two independent events, find their joint probability for the following. a. \(P(A)=.17\) and \(P(B)=.44\) b. \(P(A)=.72\) and \(P(B)=.84\)

Consider the following games with two dice. a. A gambler is going to roll a die four times. If he rolls at least one 6, you must pay him $$\$ 5 .$$ If he fails to roll a 6 in four tries, he will pay you $$\$ 5 .$$ Find the probability that you must pay the gambler. Assume that there is no cheating. b. The same gambler offers to let you roll a pair of dice 24 times. If you roll at least one double 6 , he will pay you $$\$ 10$$. If you fail to roll a double 6 in 24 tries, you will pay him $$\$ 10$$. The gambler says that you have a better chance of winning because your probability of success on each of the 24 rolls is $$1 / 36$$ and you have 24 chances. Thus, he says, your probability of winning $$\$ 10$$ is \(24(1 / 36)=2 / 3 .\) Do you agree with this analysis? If so, indicate why. If not, point out the fallacy in his argument, and then find the correct probability that you will win.

Two thousand randomly selected adults were asked whether or not they have ever shopped on the Internet. The following table gives a two-way classification of the responses. $$ \begin{array}{lcc} \hline & \text { Have Shopped } & \text { Have Never Shopped } \\ \hline \text { Male } & 500 & 700 \\ \text { Female } & 300 & 500 \\ \hline \end{array} $$ Suppose one adult is selected at random from these 2000 adults. Find the following probabilities. a. \(P(\) has never shopped on the Internet or is a female) b. \(P(\) is a male \(o r\) has shopped on the Internet) c. \(P\) (has shopped on the Internet or has never shopped on the Internet)
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