Question

An urn contains 4 red and 6 green balls. A ball is selected at random from the urn and is replaced back into the urn. Now a ball is drawn again from the bag. What is the probability that the first ball is a red ball and the second is a green ball? (1) \(\frac{6}{25}\) (2) \(\frac{8}{25}\) (3) \(\frac{5}{6}\) (4) \(\frac{4}{15}\)

Short answer

Answer: \(\frac{6}{25}\)

Step-by-step solution

Find the probability of drawing a red ball first

To find the probability of drawing a red ball first, we can use the following formula: \(P (\text{red}) = \frac{\text{number of red balls}}{\text{total number of balls}}\) There are 4 red balls, and a total of 4 + 6 = 10 balls in the urn. So, \(P (\text{red}) = \frac{4}{10} = \frac{2}{5}\)

Find the probability of drawing a green ball after replacing the first ball

Since the first ball was replaced back into the urn, we still have a total of 10 balls in the urn. We can use the same formula to find the probability of drawing a green ball: \(P (\text{green}) = \frac{\text{number of green balls}}{\text{total number of balls}}\) There are 6 green balls, so \(P (\text{green}) = \frac{6}{10} = \frac{3}{5}\)

Calculate the probability of both events happening

To find the probability of both events happening, we multiply the probabilities of the individual events: \(P (\text{red, then green}) = P (\text{red}) \times P(\text{green})\) Plugging in our values from Steps 1 and 2, we have: \(P (\text{red, then green}) = \frac{2}{5} \times \frac{3}{5} = \frac{6}{25}\) So, the probability that the first ball is a red ball and the second is a green ball is \(\frac{6}{25}\). Thus, the correct answer is (1) \(\frac{6}{25}\).

Key concepts

Conditional Probability

In probability theory, conditional probability is the probability of an event occurring given that another event has already occurred. It is expressed as \(P(A|B)\), which represents the probability of event \(A\) given that \(B\) has occurred.
In the context of the exercise, however, because the ball drawn is replaced back into the urn before drawing again, each draw is independent of the previous one, making conditional probability irrelevant here. Each event does not affect the other, unlike typical conditional probability scenarios.
The key takeaway is that when events are independent, the probability of both events occurring can be found by multiplying the probabilities of each individual event. This approach simplifies the calculation and differs from situations where events directly influence each other.

Random Experiment

A random experiment is a process or action that leads to one of several possible outcomes, where the outcome cannot be predicted with certainty. Drawing balls from an urn is an ideal example of a random experiment.
Random experiments have the following characteristics:

• Multiple possible outcomes (e.g., drawing a red ball or a green ball).
• The outcome is unknown until the experiment is conducted.
• It can be repeated under the same conditions.

In this exercise, selecting a ball from an urn where each ball has an equal chance of being chosen demonstrates these characteristics. It involves uncertainty, as we cannot foresee which ball will be picked before conducting the experiment. The random experiment is fundamental in understanding chance, likelihood, and predicting results in probability theory.

Combinatorics

Combinatorics is a branch of mathematics dealing with the counting, arrangement, and combination of objects. In probability, combinatorics is used to calculate the likelihood of various outcomes by determining potential combinations or arrangements of events.
This exercise can be linked to combinatorics by considering the selection and likelihood of combinations of balls drawn. However, since events are independent, straightforward counting of favorable outcomes is applied. The possibilities are calculating separate probabilities at each step, not considering all potential combinations.
For instance, determining the number of red and green balls involves combinatorial thinking, but no complex combinations needs to be evaluated in this particular exercise. Instead, this is one of the scenarios where simple counting (4 red and 6 green) directly translates into the fractions used to determine probability.