There are 3 red and 2 white balls in one box and 4 red and 5 white in the second box. You select a box at random and from it pick a ball at random. If the ball is red, what is the probability that it came from the second box?

Short Answer

Expert verified
The probability that the ball came from the second box given that it is red is \( \frac{20}{47} \).

Step by step solution

01

- Define Events

Let Box 1 be B1 and Box 2 be B2. Let Red Ball be R. We need to find P(B2|R), the probability that the ball came from the second box given that it is red.
02

- Use Bayes' Theorem

Bayes' theorem states: \[ P(B2|R) = \frac{P(R|B2) \, P(B2)}{P(R)} \] where, P(B1) = P(B2) = 0.5 (since you select a box at random)
03

- Calculate Probability of Selecting a Red Ball from Each Box

\[ P(R|B1) = \frac{3}{3+2} = \frac{3}{5} \] and \[ P(R|B2) = \frac{4}{4+5} = \frac{4}{9} \]
04

- Calculate the Total Probability of Selecting a Red Ball

\[ P(R) = P(R|B1) \, P(B1) + P(R|B2) \, P(B2) = \left( \frac{3}{5} \times 0.5 \right) + \left( \frac{4}{9} \times 0.5 \right) = \frac{3}{10} + \frac{2}{9} = \frac{27 + 20}{90} = \frac{47}{90} \]
05

- Final Calculation

Using Bayes' theorem, we get: \[ P(B2|R) = \frac{\left( \frac{4}{9} \times 0.5 \right)}{\frac{47}{90}} = \frac{2}{9} \times \frac{90}{47} = \frac{20}{47} \]

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Key Concepts

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

conditional probability
Conditional probability is the likelihood of an event occurring given that another event has already occurred. In our problem, we need to find the probability that a red ball came from the second box, knowing that the drawn ball is red. This is symbolized as P(B2|R). With conditional probability, you're focusing on a subset of the entire sample space. For our problem, the subset is the event where the ball is red, and we consider how likely it originated from box 2.
probability theory
Probability theory involves studying how likely events are to happen. It provides a mathematical way to measure uncertainty. Various rules and theorems within this field help to calculate probabilities.
In our problem, we use the axioms of probability to define initial probabilities like P(B1) and P(B2). Each box has equal likelihood, thus P(B1) = P(B2) = 0.5.
We also calculate specific event probabilities, such as the chance of drawing a red ball from each box: P(R|B1) = 3/5 for box 1 and P(R|B2) = 4/9 for box 2.
probabilistic reasoning
Probabilistic reasoning is the method of applying probability to make predictions or reach conclusions. It involves understanding how different probabilities interact and influence one another.
In our problem, it's used to structure the solution logically:
  • First, we identify key events: drawing from box 1 (B1) and drawing from box 2 (B2).
  • Next, we find the probability of drawing a red ball from each box.
  • We then use this information to determine the overall probability of drawing a red ball (P(R)).
This reasoning process culminates in applying Bayes' Theorem to find our desired probability.
bayesian inference
Bayesian inference involves updating our belief about a probability after obtaining new evidence.
We applied Bayes' Theorem to find the probability that the red ball came from the second box, given that it is red. Bayes' Theorem is given by:
P(B2|R) = (P(R|B2) * P(B2)) / P(R).

Applying the values, we get:
P(B2|R) = (4/9 * 0.5) / (47/90) = 20/47.
Bayesian inference thus enables us to update probabilities based on new data. It's a crucial method in probability and statistics.

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Most popular questions from this chapter

Show that the expectation of the sum of two random variables defined over the same sample space is the sum of the expectations. Hint: Let \(p_{1}, p_{2}, p_{3}, \cdots, P_{*}\) be the probabilitics associated with the \(n\) sample points; let \(x_{1}, x_{2}, \cdots, x_{n}\), and \(y_{1}, y_{2}, \cdots, y_{n}\), be the values of the random variables \(x\) and \(y\) for the \(n\) sample points. Write out \(E(x), E(y)\), and \(E(x+y)\).

A single card is drawn at random from a shuffled deck. What is the probability that it is red? That it is the ace of hearts? That it is either a three or a five? That it is either an ace or red or both?

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Two dice are thrown. Use the sample space (2.4) to answer the following questions. (a) What is the probability of being able to form a two-digit number greater than 33 with the two numbers on the dice? (Note that the sample point 1, 4 yields the two-digit number 41 which is greater than 33, etc.) (b) Repeat part (a) for the probability of being able to form a two-digit number greater than or equal to 42 (c) Can you find a two-digit number (or numbers) such that the probability of being able to form a larger number is the same as the probability of being able to form a smaller number? [See note, part (a).]

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