Question

(a) There are 3 red and 5 black balls in one box and 6 red and 4 white balls in another. If you pick a box at random, and then pick a ball from it at random, what is the probability that it is red? Black? White? That it is either red or white? (b) Suppose the first ball selected is red and is not replaced before a second ball is drawn. What is the probability that the second ball is red also? (c) If both balls are red, what is the probability that they both came from the same box?

Short answer

The probability that both balls came from either box is calculated using Bayes' theorem, though specific percentages for each box can be derived from the detailed steps.

Step-by-step solution

Define the Events

Let Event A be picking the first box and Event B be picking the second box. Define Event R for picking a red ball, Event Bl for picking a black ball, and Event W for picking a white ball. Calculate the individual probabilities of these events.

Determine Box Probabilities

Calculate the probability of picking each box. Since each box is equally likely, $P(A)=P(B)=\frac{1}{2}$.

Determine Red Ball Probabilities

Calculate the probability of picking a red ball from each box. For the first box: $P(R|A)=\frac{3}{8}$. For the second box: $P(R|B)=\frac{6}{10}=\frac{3}{5}$.

Calculate Total Probability for Red Ball

Use the Law of Total Probability to calculate the overall probability of picking a red ball: $$P(R)=P(A)\cdot P(R|A)+P(B)\cdot P(R|B)=\frac{1}{2}\cdot \frac{3}{8}+\frac{1}{2}\cdot \frac{3}{5}=\frac{3}{16}+\frac{3}{10}=\frac{15}{80}+\frac{24}{80}=\frac{39}{80}.$$

Calculate Black and White Ball Probabilities

For black balls, only the first box has them: $$P(Bl)=P(A)\cdot P(Bl|A)+P(B)\cdot P(Bl|B)=\frac{1}{2}\cdot \frac{5}{8}+\frac{1}{2}\cdot 0=\frac{5}{16}.$$ For white balls, only the second box has them: $$P(W)=P(A)\cdot P(W|A)+P(B)\cdot P(W|B)=\frac{1}{2}\cdot 0+\frac{1}{2}\cdot \frac{4}{10}=\frac{2}{10}=\frac{1}{5}.$$

Calculate Probability of Red or White

Calculate the probability of picking a red or white ball: $$P(R\cup W)=P(R)+P(W)=\frac{39}{80}+\frac{1}{5}=\frac{39}{80}+\frac{16}{80}=\frac{55}{80}=\frac{11}{16}.$$

Calculate Probability for the Second Red Ball

For the first box, if the first red ball is removed: $$P(R_2|R_1)=\frac{2}{7}$$ For the second box: $$P(R_2|R_1)=\frac{5}{9}.$$

Calculate Probability both Red Balls from the Same Box

Given both balls are red, use Bayes theorem: $$P(A|R_1\cap R_2)=\frac{P(A\cap R_1\cap R_2)}{P(R_1\cap R_2)}$$ and $$P(B|R_1\cap R_2)=\frac{P(B\cap R_1\cap R_2)}{P(R_1\cap R_2)}.$$

Key concepts

Conditional Probability

Conditional probability is the likelihood of an event occurring given that another event has already occurred. In our problem, one good example is finding the probability of drawing a second red ball given that the first ball drawn was red. This scenario is described by the notation:

$P(R_2|R_1)$, which means the probability that the second ball is red given the first ball was red. The formula is straightforward:

$$P(R_2|R_1)=\frac{P(R_2\text{ and }{R}_1)}{P(R_1)}$$.

For example, after removing one red ball from the first box, the probability of drawing another red ball from it is $\frac{2}{7}$. Similarly, from the second box, it is $\frac{5}{9}$. This factor depends on the first event (drawing a red ball). Clearly, conditional probabilities help us capture the essence of how one event influences another.

Law of Total Probability

The Law of Total Probability helps us find the overall probability of an event by considering all possible scenarios that could occur. In our exercise, this law allows us to calculate the total probability of drawing a red ball from either of the two boxes.

The formula states:

$$P(R)=P(A)\times P(R|A)+P(B)\times P(R|B)$$.

Here, $P(A)$ and $P(B)$ are the probabilities of picking the first and the second box respectively, both equal to $\frac{1}{2}$. $P(R|A)$ and $P(R|B)$ are the conditional probabilities of drawing a red ball from the first and second boxes. When applying the law, we get:

$$P(R)=\frac{1}{2}\times \frac{3}{8}+\frac{1}{2}\times \frac{3}{5}=\frac{39}{80}$$.

This computation gives us the comprehensive likelihood of drawing a red ball by accounting for both boxes.

Bayes Theorem

Bayes Theorem allows us to revise probabilities given new information. It is particularly useful when we need to update our probabilities after observing some event.

In the given exercise, part (c) asks us to find the probability that both red balls came from the same box, considering they were already drawn red. Using Bayes Theorem, this can be computed as:

$$P(A|R_1\text{ and }{R}_2)=\frac{P(A\text{ and }{R}_1\text{ and }{R}_2)}{P(R_1\text{ and }{R}_2)}$$.

If $P(R_1\text{ and }{R}_2)$ is the joint probability of both red balls, and $P(A\text{ and }{R}_1\text{ and }{R}_2)$ is the probability both red balls coming from the first box, we can simplify to:

$$P(A|R_1\text{ and }{R}_2)=\frac{P(A)\times P(R_1\text{ and }{R}_2|A)}{P(R_1\text{ and }{R}_2)}$$.

Bayes theorem thus connects our probabilities before and after observing the two red balls, guiding us in updating likelihoods effectively.

Event Definition

Events define possible outcomes or occurrences in a probability scenario. In this exercise, multiple events are defined:

• Event A: Picking the first box
• Event B: Picking the second box
• Event R: Drawing a red ball
• Event Bl: Drawing a black ball
• Event W: Drawing a white ball

The probability of these events guides us in solving the problem. By calculating the probability of individual events and their intersections, we can break down complex problems into smaller, manageable parts. For instance, to find the probability of drawing a black ball $P(Bl)$, we only consider that box one has black balls.

Combinatorial Probability

Combinatorial Probability focuses on counting the number of ways an event can occur relative to the total number of possible outcomes. In our exercise, combinatorial methods help analyze scenarios such as removing a ball from the boxes without replacement.

To compute the probability of drawing another red ball after one has already been removed, we use combinations:

From the first box, if the first red ball is removed, the ways to draw another red are reduced: $$P(R_2|R_1)=\frac{2}{7}$$.

From the second box, the combinations are similarly adjusted: $$P(R_2|R_1)=\frac{5}{9}$$.

These outcomes rely on preventing repetition, aligning well with combinatorial principles.