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Chapter 9: Problem 38

A jar contains four blue marbles and two red marbles. Suppose you choose a marble at random, and do not replace it. Then you choose a second marble. Find the probability of each event. One of the marbles you select is blue and the other is red.

Short Answer

Expert verified
The probability that one of the marbles selected is blue and the other is red is \(8/15\).

Step by step solution

01

Understand the Problem

The jar contains 4 blue and 2 red marbles, making it a total of 6 marbles. Two marbles are selected consecutively without replacement. The task here is to find the probability that one of the marbles is red and the other one is blue.
02

Calculate the Probability of Selecting one Blue and one Red Marble

There are two sequence of events that can lead to the selection of one red and one blue marble, either a red followed by a blue or blue followed by a red marble is selected. These are two distinct events in probability.\n\nFirstly, if a blue marble is selected first, the probability is 4 out of 6 (since there are 4 blue marbles out of a total of 6): \(P(B1)= 4/6 = 2/3\). After the first draw, there are now 5 marbles left, hence the probability of selecting a red marble next is 2 out of 5 : \(P(R2|B1) = 2/5\). \n\nAlternatively, if a red marble is selected first, the probability is 2 out of 6: \(P(R1)= 2/6 = 1/3\). Then, the probability of choosing a blue marble from the 5 remaining is 4 out of 5: \(P(B2|R1) = 4/5\).
03

Apply the Multiplication Rule

The multiplication rule of probability states that the probability of occurrence of two dependent events is the product of the probability of the first event and the conditional probability of the second event.\n\nThe conditional probability for the first sequence of events (Blue then Red) is \( P(B1 and R2) = P(B1) * P(R2|B1) = 2/3 * 2/5 = 4/15 \).\n\nThe conditional probability for the second sequence of events (Red then Blue) is \( P(R1 and B2) = P(R1) * P(B2|R1) = 1/3 * 4/5 =4/15\).
04

Sum the Probabilities of the Two distinct Events: 'Blue then Red' or 'Red then Blue'

The two sequences of events: 'Blue then Red' and 'Red then Blue' considered above are distinct, hence, to get the overall probability, their probabilities should be added according to the addition rule of probability: \( P = P(B1 and R2) + P(R1 and B2) = 4/15 + 4/15 = 8/15 \)

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Key Concepts

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

Probability without replacement
When calculating the probability of events occurring without replacement, you're dealing with a scenario where each item, once chosen, isn't returned to the original pool. This affects the outcome of subsequent draws. In our example, a marble is picked from a jar containing both blue and red marbles, but once a marble is chosen, it鈥檚 not placed back.
This changes the total number of marbles available for the next draw, impacting the probabilities. For instance, after removing one marble, there are now 5 marbles remaining instead of 6. Removing an item alters the sample space and, consequently, the odds of future draws.
Dependent events
Dependent events are events where the occurrence of one affects the probability of the other. Unlike independent events, dependent events are closely linked.
In the context of our marble problem, after selecting the first marble, the color of that marble influences what options are available next. If a blue marble is picked first, the likelihood of picking any other color changes because the pool of available marbles has altered. This dependency defines how each subsequent event must be recalculated based on the results of prior draws.
Conditional probability
Conditional probability explores the likelihood of an event occurring, given that another event has already happened. It鈥檚 denoted as \(P(A|B)\), meaning the probability of event A occurring given B has occurred.
In our marble example, if the first draw results in a blue marble, the probability of selecting a red marble next becomes a conditional probability, affected by the outcome of the first event. Initially, there might be 2 red marbles out of 6, but after removing a blue, we're left with 2 reds out of 5. This illustrates how conditional probability recalibrates expectations based on new circumstances.
Multiplication rule of probability
The multiplication rule is pivotal for finding the probability of two dependent events occurring consecutively. The rule states that for two events, A and B, the joint probability \(P(A \text{ and } B)\) is \(P(A)\) times \( P(B|A)\), the conditional probability of B given A.
Applied to our scenario, if you first draw a blue marble, then the conditional probability of drawing a red marble remains contingent upon the first outcome. Thus, you calculate \(P(B1) \cdot P(R2|B1)\) or the reverse \( P(R1) \cdot P(B2|R1) \) and sum these results for the total probability. This method helps ascertain the likelihood of consecutive, dependent events by considering the outcomes sequentially.

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