Problem 19
Question
A drawer contains three bags numbered \(1-3\), respectively. Bag 1 contains three blue balls, bag 2 contains four green balls. and bag 3 contains two blue balls and one green ball. You choose one bag at random and take out one ball. Find the probability that the ball is blue.
Step-by-Step Solution
Verified Answer
The probability of picking a blue ball is \( \frac{5}{9} \).
1Step 1: Determine Total Outcomes
There are three bags, and each has a chance of being chosen. Thus, there are three possible outcomes, one for each bag.
2Step 2: Calculate Probability of Each Bag
Since the choice of the bag is random, the probability of choosing any specific bag is equal. Therefore, the probability for each bag is \( \frac{1}{3} \).
3Step 3: Find Probability of Picking a Blue Ball from Each Bag
- **Bag 1:** It has 3 blue balls and no other colors, so \( P(\text{blue} | \text{Bag 1}) = 1 \).- **Bag 2:** It has no blue balls, so \( P(\text{blue} | \text{Bag 2}) = 0 \).- **Bag 3:** It has 2 blue balls out of 3, so \( P(\text{blue} | \text{Bag 3}) = \frac{2}{3} \).
4Step 4: Use Total Probability Theorem
The probability of drawing a blue ball is given by:\[P(\text{blue ball}) = P(\text{Bag 1}) \cdot P(\text{blue} | \text{Bag 1}) + P(\text{Bag 2}) \cdot P(\text{blue} | \text{Bag 2}) + P(\text{Bag 3}) \cdot P(\text{blue} | \text{Bag 3})\]Substitute the known probabilities:\[P(\text{blue ball}) = \frac{1}{3} \times 1 + \frac{1}{3} \times 0 + \frac{1}{3} \times \frac{2}{3}\]
5Step 5: Perform the Calculations
Calculating the expression:\[P(\text{blue ball}) = \frac{1}{3} + 0 + \frac{2}{9} = \frac{3}{9} + \frac{2}{9} = \frac{5}{9}\]
6Step 6: Conclude the Probability
The probability of picking a blue ball is \( \frac{5}{9} \).
Key Concepts
Conditional ProbabilityTotal Probability TheoremCombinatorics
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already occurred. In this exercise, we're trying to find the probability of drawing a blue ball from one of the bags given that we chose a bag first. The formula for conditional probability is:
- \(P(A | B) = \frac{P(A \cap B)}{P(B)}\)
- \( A \) is drawing a blue ball
- \( B \) involves choosing a specific bag
Total Probability Theorem
The total probability theorem allows us to find the probability of an event by considering all different paths that lead to it, weighted by the likelihood of those paths. The formula is:
- \(P(A) = \sum P(B_i) \cdot P(A | B_i)\)
- \( A \) is the event of interest (e.g., drawing a blue ball)
- \( B_i \) are mutually exclusive events (e.g., choosing each individual bag)
- From Bag 1: \(\frac{1}{3} \times 1\)
- From Bag 2: \(\frac{1}{3} \times 0\)
- From Bag 3: \(\frac{1}{3} \times \frac{2}{3}\)
- \(P(\text{blue ball}) = \frac{1}{3} + 0 + \frac{2}{9} = \frac{5}{9}\)
Combinatorics
Combinatorics is the area of mathematics that deals with counting, arranging, and structuring elements within a set. In our problem, although we focused on probability, understanding the combinations of balls in each bag is key.
Combinatorial analysis would typically involve exploring different ways in which combinations can occur. However, in this exercise, it's fairly straightforward since each bag's content is directly given, eliminating the need for complex calculations.
Important points to consider in combinatorial problems include:
- The total number of outcomes possible
- The probability of selecting certain combinations
- How distinct scenarios interact (e.g., different bag choices)
Other exercises in this chapter
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