Problem 7 - A First Course in Probability

Question

Suppose that \(A\) and \(B\) each randomly and independently choose 3 of 10 objects. Find the expected number of objects (a) chosen by both \(A\) and \(B\) (b) not chosen by either \(A\) or \(B\) (c) chosen by exactly one of \(A\) and \(B\)

Step-by-Step Solution

Answer

The expected number of objects (a) chosen by both A and B is \(\frac{9}{10}\), (b) not chosen by either A or B is \(\frac{49}{10}\), and (c) chosen by exactly one of A and B is \(\frac{42}{10}\).

Step 1: a) Expected number of objects chosen by both A and B

We have to find the expected value of objects chosen by both A and B. Let's take an object, say x. Now, we will find the probability that both A and B choose x. The probability that A chooses x is: \(P_A = \frac{3}{10}\) The probability that B chooses x is: \(P_B = \frac{3}{10}\) Since both choices are independent, the probability that both A and B choose x = \(P_{AB} = P_A * P_B =\frac{3}{10} * \frac{3}{10} = \frac{9}{100}\) Now, there are 10 objects in total. So the expected number of objects chosen by both A and B is the sum of the probabilities of each object being chosen by both A and B. The expected number of objects chosen by both A and B = \(10* P_{AB} = 10 * \frac{9}{100} = \frac{9}{10}\)

Step 2: b) Expected number of objects not chosen by either A or B

Let's find the probability that neither A nor B chooses object x. The probability that A does not choose x is: \(P_{A'} = 1-P_A = \frac{7}{10}\) The probability that B does not choose x is: \(P_{B'} = 1-P_B = \frac{7}{10}\) Since A and B choices are independent, The probability that both A and B do not choose x = \(P_{A'B'} = P_{A'} * P_{B'} =\frac{7}{10} * \frac{7}{10} = \frac{49}{100}\) Now, we have 10 objects in total. The expected number of objects not chosen by either A or B is the sum of the probabilities of each object not being chosen by A and B. The expected number of objects not chosen by either A or B = \(10 * P_{A'B'}= 10* \frac{49}{100} = \frac{49}{10}\)

Step 3: c) Expected number of objects chosen by exactly one of A and B

We have calculated the expected number of objects chosen by both A and B, and not chosen by either A or B. Now let's find the expected number of objects chosen by exactly one of them, either A or B. Using the property of expected values, the expected number of objects chosen by exactly one of A and B = The total number of objects - the expected number of objects chosen by both A and B - the expected number of objects not chosen by either A or B. The expected number of objects chosen by exactly one of A and B = \(10 - \frac{9}{10} - \frac{49}{10} = 10 - \frac{58}{10} = \frac{42}{10}\) So the expected number of objects chosen by exactly one of A and B is \(\frac{42}{10}\).

Key Concepts

Independent Events

When you're dealing with independent events in probability, it's like flipping a coin or rolling a die where one action doesn't affect the other. Imagine you're choosing objects without any "peer pressure." Here, we're given two events: choosing objects by individuals A and B. Their decisions are made independently, without looking over each other's shoulders.
This means that if A chooses an object, it doesn't affect B's choice, or vice versa. So, the probability of A or B choosing an object is determined solely by their own independent actions.

• A's choice of an object: Probability is \(\frac{3}{10}\) because A chooses 3 out of 10 objects.
• B's choice of an object: Similarly, the probability is \(\frac{3}{10}\).

Combining these independent actions means we multiply their probabilities to find the chance of both events occurring at the same time. For example, the probability of both A and B choosing the same object is \(\frac{3}{10} \times \frac{3}{10} = \frac{9}{100}\).
This shows how independence in probability allows you to handle each event separately when calculating their joint probabilities.

Probability Calculation

Probability calculation can be thought of as a straightforward arithmetic of chance. For any event, we calculate probability based on the ratio of favorable outcomes to all possible outcomes. Let's look at specific scenarios from the problem to understand how this works in practice.
First off, consider how we calculate the probability of neither A nor B selecting a particular object. For this, you need to find out how likely it is that both won't pick the object.

• The probability of A not picking the object: \(P_{A'} = 1 - \frac{3}{10} = \frac{7}{10}\).
• The probability of B not picking the object: \(P_{B'} = 1 - \frac{3}{10} = \frac{7}{10}\).
• Since A and B act independently, the combined probability of neither choosing it is:\(P_{A'B'} = \frac{7}{10} \times \frac{7}{10} = \frac{49}{100}\).

Once the probability is calculated, extending this to all objects gives us the expected count across multiple trials. Multiply the probability by the total number of objects which, in this case, results in \(\frac{49}{10}\) objects not being chosen by either.
Understanding these probabilities lets you confidently tackle related problems by applying basic principles to compute expectations.

Combinatorial Probability

Combinatorial probability relates to calculating probabilities where you choose items from a larger set, often dealing with concepts like the "binomial coefficient." Here, the idea is subtly present in calculating expectations for numbers of objects chosen or not chosen between A and B.The expected value, effectively an "average outcome," is particularly helpful when addressing combinatorial aspects. Consider how many objects are expected to be chosen by exactly one of A or B. We calculated it as:

• Total objects: 10
• Subtract those chosen by both (\(\frac{9}{10}\)) and those by neither (\(\frac{49}{10}\)) from the total:
• The equation simplifies to: \(10 - \frac{9}{10} - \frac{49}{10} = \frac{42}{10}\). This gives the expected number chosen by only one, showing an effective use of subtraction in combinatorial setups.

Though less explicit than other exercises, combinatorial probability here revolves around effective organization and synthesis of probabilities to understand larger systems. It underscores the versatility of combinatorial arrangements, even in straightforward problems like choosing objects independently.