Probability of Selecting Two Red Marbles from a Bowl Without Replacement

In this article, we will explore the probability of selecting two red marbles from a bowl containing 7 green and 3 red marbles without replacement. We will break down the steps required to calculate this probability and understand the underlying concepts.

Understanding the Problem

The problem involves determining the probability of drawing two red marbles consecutively without replacing the first marble. This is a classic example of probability calculation where the sample space changes after each draw. Let's dive into the detailed steps to solve this problem.

Step-by-Step Solution

Determine the Total Number of Marbles

Let's start by identifying the total number of marbles in the bowl.

Calculate the Total Ways to Choose 2 Marbles from 10

We will use the combination formula to determine the total number of ways to choose 2 marbles from 10 without regard to order.

The combination formula is given by: C(n, k) (frac{n!}{k!(n-k)!})

For our problem, C(10, 2) (frac{10!}{2!(10-2)!}) (frac{10 times 9}{2 times 1}) 45

Calculate the Ways to Choose 2 Red Marbles from 3

Next, we need to determine the number of ways to choose 2 red marbles from the 3 available red marbles.

Using the combination formula again, we get:

C(3, 2) (frac{3!}{2!(3-2)!}) (frac{3 times 2}{2 times 1}) 3

Calculate the Probability

The probability of both marbles being red is the number of favorable outcomes divided by the total number of possible outcomes.

P(Both Red) (frac{3}{45}) (frac{1}{15})

Therefore, the probability that both marbles selected are red is (frac{1}{15}).

Different Perspectives on the Problem

The problem can also be approached from the perspective of drawing two marbles in a specific order, with the first marble being green and the second being red.

Probability of First Ball Being Green and Second Ball Being Red

First, calculate the probability of drawing a green ball first:

P(Green First) (frac{7}{12})

After drawing a green ball, there are now 11 marbles left, of which 3 are red. The probability of drawing a red ball second is:

P(Red Second | Green First) (frac{3}{11})

The combined probability of both events happening (Green First and Red Second) is the product of the individual probabilities:

P(Green First (cap) Red Second) (frac{7}{12} times frac{3}{11} frac{35}{132})

The probability that the first ball selected is green and the second ball selected is red is (frac{35}{132}).

Summary

In this article, we have explored the problem of calculating the probability of selecting two red marbles from a bowl without replacement. We discussed both the combinatorial approach and the sequential probability approach. The key concepts involved are the combination formula and the rule of multiplication for independent events.

Frequently Asked Questions

What is the probability of drawing two marbles without replacement?

The probability of drawing two marbles without replacement is based on the concept of probability and combinatorics. It involves calculating the number of favorable outcomes and dividing it by the total number of possible outcomes.

How does the combination formula work?

The combination formula, denoted as C(n, k), is used to determine the number of ways to choose k items from a set of n items without regard to order. It is given by C(n, k) (frac{n!}{k!(n-k)!}).

What is the difference between independent and dependent events?

Independent events are those where the outcome of one event does not affect the probability of the other. Dependent events, such as drawing marbles without replacement, are influenced by previous outcomes.