Question

A box contains four black pieces of cloth, two striped pieces, and six dotted pieces. A piece is selected randomly and then placed back in the box. A second piece is selected randomly. What is the probability that: a. both pieces are dotted? b. the first piece is black and the second piece is dotted? c. one piece is black and one piece is striped?

Short answer

a. \(\frac{1}{4}\), b. \(\frac{1}{6}\), c. \(\frac{1}{9}\)

Step-by-step solution

Identify the Total Number of Pieces

First, calculate the total number of pieces in the box. There are four black pieces, two striped pieces, and six dotted pieces. Thus, the total number of pieces is: \(4 + 2 + 6 = 12\).

Calculate Probability for Part (a)

For part a, we need to find the probability that both pieces are dotted and each piece is replaced before the next is selected. The probability of selecting a dotted piece on the first draw is \( \frac{6}{12} = \frac{1}{2} \). Because the piece is replaced, the probability of also selecting a dotted piece on the second draw is again \( \frac{1}{2} \). Hence, the probability that both pieces are dotted is \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).

Calculate Probability for Part (b)

For part b, we need to find the probability that the first piece is black and the second piece is dotted. The probability of selecting a black piece on the first draw is \( \frac{4}{12} = \frac{1}{3} \). Since the piece is replaced, the probability of selecting a dotted piece on the second draw is \( \frac{6}{12} = \frac{1}{2} \). Hence, the probability that the first piece is black and the second is dotted is \( \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \).

Calculate Probability for Part (c)

For part c, we need to find the probability that one piece is black and one piece is striped, and either order is possible. First, calculate the probability for one specific order (black first, striped second). The probability of selecting a black piece first is \( \frac{4}{12} = \frac{1}{3} \) and, with replacement, the probability of selecting a striped piece second is \( \frac{2}{12} = \frac{1}{6} \). So, the probability for this order is \( \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} \). Repeat the calculation for the order striped first, black second. The probability of selecting a striped piece first is \( \frac{2}{12} = \frac{1}{6} \) and for a black piece second is \( \frac{4}{12} = \frac{1}{3} \). This gives the same probability \( \frac{1}{6} \times \frac{1}{3} = \frac{1}{18} \). Adding both orders gives \( \frac{1}{18} + \frac{1}{18} = \frac{2}{18} = \frac{1}{9} \).

Key concepts

Conditional Probability

Conditional probability is a crucial concept in probability theory. It refers to the probability of an event occurring given that another event has already occurred. Understanding it helps us make predictions in scenarios where outcomes are interdependent.
Imagine needing to find the probability of a second event, knowing what happened in the first. The formula for conditional probability is: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] where \( P(A|B) \) is the probability of event A occurring given that B has happened, and \( P(A \cap B) \) is the probability that both A and B occur. In the context of the exercise, when dealing with replacement, each draw is independent. Hence, the concept of conditional probability doesn't apply directly to this exercise's solutions, as each selection is not affected by the previous one.

Independent Events

Independent events are events where the probability of one occurring doesn't affect the other. Whenever you have such a scenario, you multiply the probabilities of each event to find the probability of both events happening together.
For instance, if you have an event A and an event B, and they are independent, the probability of both A and B occurring is: \[ P(A \cap B) = P(A) \times P(B) \] In our exercise, each piece drawn from the box is replaced before the next draw. This replacement ensures the events (each draw) remain independent. For example, calculating the probability of selecting two dotted pieces involves this principle since each dotted piece draw is independent: \[ \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \] Such thinking is key when dealing with situations like this one, where replacement keeps the probabilities consistent.

Combinatorial Probability

Combinatorial probability involves counting the number of ways certain outcomes can happen and is useful in probabilities of sequential events where you order selections or count combinations.
In combinatorial probability, you use principles like permutations, combinations, and factorials to solve complex problems, especially when the order of selection is important as in calculating different ways to arrange or select objects.
In our exercise's part c, calculating the probability that one piece is black and one is striped involves considering two possible orders: black first or striped first. You account for both orderings by performing calculations separately and then adding them to find the total probability:

• Black first then striped: \( \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} \)
• Striped first then black: \( \frac{1}{6} \times \frac{1}{3} = \frac{1}{18} \)

Adding these results gives a total probability of \( \frac{1}{9} \). Combinatorial methods help solve such problems by considering all valid permutations or combinations of outcomes.