Problem 25 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Five black balls and four white balls are placed in an urn. Two balls are then drawn in succession. What is the probability that the second ball drawn is a white ball if a. The second ball is drawn without replacing the first? b. The first ball is replaced before the second is drawn?

Step-by-Step Solution

Answer

The probability of drawing a white ball as the second ball is: a. Without replacement: \(\frac{4}{9}\) b. With replacement: \(\frac{4}{9}\)

Step 1: Part a: Probability without Replacement

In this scenario, we first find the probability of drawing a white ball as the second ball given that a black ball was drawn first, and then given that a white ball was drawn first. We will use the formula for conditional probability: \[P(B_2|A) = \frac{P(B_2 \cap A)}{P(A)}\] where \(B_2\) is the event of drawing a white ball as the second ball, and \(A\) is the event related to the first ball drawn. 1. If a black ball is drawn first, - \(P(\)black ball drawn first\) = \frac{5}{9}\) - \(P(\)white ball drawn as second\ |\ black ball drawn first\) = \frac{4}{8} = \frac{1}{2}\), since there are now 8 balls left in the urn. - \(P(\)white ball drawn as second \cap black ball drawn first\) = \frac{5}{9} \times \frac{1}{2} = \frac{5}{18}\) 2. If a white ball is drawn first, - \(P(\)white ball drawn first\) = \frac{4}{9}\) - \(P(\)white ball drawn as second\ |\ white ball drawn first\) = \frac{3}{8}\), since there are now 8 balls left in the urn. - \(P(\)white ball drawn as second \cap white ball drawn first\) = \frac{4}{9} \times \frac{3}{8} = \frac{1}{6}\) We now sum the probabilities from both cases to get the overall probability: \(P(\)white ball as second ball without replacement\) = \frac{5}{18} + \frac{1}{6} = \frac{5 + 3}{18} = \frac{8}{18} = \frac{4}{9}\)

Step 2: Part b: Probability with Replacement

In this scenario, we follow the same steps as in part a, but the probabilities change since the first ball is replaced before the second is drawn. 1. If a black ball is drawn first, - \(P(\)black ball drawn first\) = \frac{5}{9}\) - \(P(\)white ball drawn as second\ |\ black ball drawn first\) = \frac{4}{9}\) (no change in the number of balls since replacement). - \(P(\)white ball drawn as second \cap black ball drawn first\) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}\) 2. If a white ball is drawn first, - \(P(\)white ball drawn first\) = \frac{4}{9}\) - \(P(\)white ball drawn as second\ |\ white ball drawn first\) = \frac{4}{9}\) (no change in the number of balls since replacement). - \(P(\)white ball drawn as second \cap white ball drawn first\) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}\) We now sum the probabilities from both cases to get the overall probability: \(P(\)white ball as second ball with replacement\) = \frac{20}{81} + \frac{16}{81} = \frac{36}{81} = \frac{4}{9}\)

Step 3: Final Answer

The probability of drawing a white ball as the second ball is: a. Without replacement: \(\frac{4}{9}\) b. With replacement: \(\frac{4}{9}\)

Key Concepts

Conditional Probability

When solving probability problems, it's often necessary to calculate the chances of an event occurring given that another event has already taken place. This is known as conditional probability. In our urn example, we're interested in the likelihood of drawing a white ball second, under the condition that we know the outcome of the first draw.

Using the formula for conditional probability, \[P(B_2|A) = \frac{P(B_2 \cap A)}{P(A)}\]where \(B_2\) is the event of drawing a white ball second, and \(A\) is the first draw event, we can determine the dependent outcomes. Understanding how to apply this formula is key to not just solving urn problems, but also to addressing a wide range of real-world scenarios where outcomes are interdependent.

Probability without Replacement

In probability theory, without replacement refers to the scenario where once an item is selected, it's not put back into the pool before the next selection. This affects subsequent probabilities since the total number of items is reduced. In our urn example, if a black ball is drawn first, one less black ball is left for future draws.

• If a black ball is drawn first (probability of \(\frac{5}{9}\)), the probability of then drawing a white ball is \(\frac{4}{8}\) because one less black ball is in the urn.
• If a white ball is drawn first (probability of \(\frac{4}{9}\)), the probability of another white ball being drawn is \(\frac{3}{8}\).

Combining these, considering the reduced number of balls after the first draw, will provide the total probability of drawing a white ball second.

Probability with Replacement

Conversely, with replacement assumes that every item selected is put back into the pool before subsequent selections, ensuring the total number of items remains constant. This means each draw is independent of the previous draws. In the urn problem, it simplifies calculations as the probabilities for the second draw remain the same as if it were the first draw.

• In both scenarios of drawing a black or white ball first, the subsequent probability of drawing a white ball remains \(\frac{4}{9}\).

When instructing on this concept, emphasize that with replacement problems assume independence, a concept vital in probability and statistics. Understanding the distinction between independent (with replacement) and dependent (without replacement) events is crucial for students tackling a wide array of probability problems.