Problem 5 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

A ball is drawn at random from an urn containing one red and one white ball. If the white ball is drawn, it is put back into the urn. If the red ball is drawn, it is returned to the urn together with two more red balls. Then a second draw is made. What is the probability a red ball was drawn on both the first and the second draws?

Step-by-Step Solution

Answer

The probability of drawing a red ball on both draws is \( \frac{3}{8} \).

Step 1: Determine Initial Probabilities

Initially, the urn contains one red ball and one white ball, so the probability of drawing a red ball first is \( P(R_1) = \frac{1}{2} \) and the probability of drawing a white ball first is \( P(W_1) = \frac{1}{2} \).

Step 2: Calculate Outcome After Drawing White Ball

If the white ball is drawn first, it is put back into the urn. The urn still contains one red and one white ball, so the probability of drawing a red ball on the second draw given the white ball was drawn first is \( P(R_2 | W_1) = \frac{1}{2} \).

Step 3: Calculate Outcome After Drawing Red Ball

If the red ball is drawn first, it's returned along with two additional red balls. The urn then contains three red balls and one white ball, so the probability of drawing a red ball on the second draw given the red ball was drawn first is \( P(R_2 | R_1) = \frac{3}{4} \).

Step 4: Calculate Joint Probability for Two Red Balls

To find the probability of drawing a red ball on both the first and the second draws, we calculate the joint probability: \( P(R_1 \text{ and } R_2) = P(R_1) \times P(R_2 | R_1) = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8} \).

Key Concepts

Conditional Probability

Conditional probability is all about determining the likelihood of an event given that another event has already occurred. In our urn example, we're interested in the probability of drawing a red ball on the second draw given that a certain outcome occurred on the first draw.
Since the conditions of the problem change depending on the first draw, the probability for the second draw will differ.
For instance, if we draw a white ball first and put it back, we are left with the same probabilities for the second draw as the first: 1 red and 1 white ball, hence a conditional probability of \( P(R_2 | W_1) = \frac{1}{2} \).
Alternatively, if the first draw yields a red ball and two additional reds are added, we see the conditional probability for drawing a red ball change due to the change in composition of the urn, leading to \( P(R_2 | R_1) = \frac{3}{4} \). Knowing the event that precedes can significantly alter the outcome of the subsequent event.

• Remember: Conditional probability is represented as \( P(A | B) \), meaning "The probability of A given B has occurred."
• The pre-condition affects the setting or context of the subsequent event.

Random Variables

In probability theory, a random variable is a variable whose possible values are numerical outcomes of a random process. It represents the outcomes of our draw in the urn problem.
Consider the influence of random variables in the urn setup; our first draw outcome could be represented by a random variable, say \( X \), which might yield either "Red" or "White" with its own distribution probabilities.
In this specific exercise, you can think of the first draw as a discrete random variable which has outcomes of drawing either a red or white ball, both initially having equal chance \( \frac{1}{2} \). The property of these variables is that they reflect the probability of all possible outcomes in a random experiment.

• Random variables have defined probability distributions that capture the likelihood of various outcomes.
• In our example's case, adding two more red balls adjusts the probability distribution for the second draw based on the first random variable's outcome.
• Understanding random variables helps to better calculate and predict total probabilities over sequences of events.

Joint Probability

Joint probability deals with the likelihood of two events occurring simultaneously. Calculate it by multiplying marginal probability of the first event by the conditional probability of the second event.
In our urn example, the joint probability refers to the probability that we draw a red ball on the first draw \( R_1 \) and also on the second draw \( R_2 \).
This is computed using the formula for joint probability: \( P(A \text{ and } B) = P(A) \times P(B | A) \). For our specific question, this becomes \( P(R_1 \text{ and } R_2) = P(R_1) \times P(R_2 | R_1) \).
Inserting the probabilities we calculated into the joint probability formula: \[ P(R_1 \text{ and } R_2) = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8} \].

• When two events are dependent, as in our scenario, the joint probability demonstrates the combined probability of both events occurring.
• Always consider the sequence in which events occur; earlier events influence later outcomes in conditional dependencies.
• This approach allows for more detailed understanding when multiple steps involve probabilistic decisions.