Probability Calculation of Marble Selection in a Bag

Let's dive into the fascinating world of probability with a practical example. Imagine you have a bag containing 3 red, 4 yellow, and 5 blue marbles. The challenge is to calculate the probabilities of various scenarios when you randomly select two marbles without replacement. Here, we will systematically work through each scenario and present the solution step-by-step.

Problem Overview

The total number of marbles in the bag is 3 red, 4 yellow, and 5 blue, totaling 12 marbles. We will use combinations to determine the probability of selecting marbles under different conditions.

Total Number of Marbles

Total Marbles in the Bag: Red: 3 Yellow: 4 Blue: 5 Total 12 marbles

Total Ways to Choose 2 Marbles

The total number of ways to choose 2 marbles out of 12 is given by the combination formula:

( binom{12}{2} frac{12!}{2!(12-2)!} 66 )

This formula, (binom{n}{k} frac{n!}{k!(n-k)!}), accounts for the number of ways to choose 2 marbles out of 12 without regard to order.

Scenario A: Both Marbles are Blue

To find the probability that both marbles are blue, we need to calculate the number of ways to choose 2 blue marbles and divide it by the total number of ways to choose 2 marbles from 12.

Number of ways to choose 2 blue marbles out of 5:

( binom{5}{2} frac{5!}{2!(5-2)!} 10 )

Probability that both marbles are blue:

( P(text{Both Blue}) frac{10}{66} frac{5}{33} )

Scenario B: Both Marbles Have the Same Color

To find the probability that both marbles have the same color, we sum the probabilities for each color.

Number of ways to choose 2 red marbles:

( binom{3}{2} frac{3!}{2!(3-2)!} 3 )

Number of ways to choose 2 yellow marbles:

( binom{4}{2} frac{4!}{2!(4-2)!} 6 )

Number of ways to choose 2 blue marbles:

( binom{5}{2} frac{5!}{2!(5-2)!} 10 )

Total number of ways to choose 2 marbles of the same color:

( 3 6 10 19 )

Probability that both marbles have the same color:

( P(text{Same Color}) frac{19}{66} )

Scenario C: At Least One Marble is Red

This scenario is easier to solve using the complement rule. We calculate the probability that neither marble is red and subtract it from 1.

Number of marbles that are not red (yellow and blue):

( 4 yellow 5 blue 9 )

Number of ways to choose 2 marbles out of 9 non-red marbles:

( binom{9}{2} frac{9!}{2!(9-2)!} 36 )

Probability that neither marble is red:

( P(text{Neither Red}) frac{36}{66} frac{6}{11} )

Probability that at least one marble is red:

( P(text{At Least One Red}) 1 - P(text{Neither Red}) 1 - frac{6}{11} frac{5}{11} )

Scenario D: Exactly One Marble is Yellow

To find the probability that exactly one marble is yellow, we can have one yellow and one non-yellow (either red or blue).

Number of ways to choose 1 yellow marble:

( binom{4}{1} frac{4!}{1!(4-1)!} 4 )

Number of ways to choose 1 non-yellow red or blue marble:

( 3 red 5 blue 8 )

Number of ways to choose 1 non-yellow marble:

( binom{8}{1} 8 )

Total ways to choose exactly one yellow and one non-yellow marble:

( 4 times 8 32 )

Probability that exactly one marble is yellow:

( P(text{Exactly One Yellow}) frac{32}{66} frac{16}{33} )

Summary of Probabilities

Probability that both are blue: ( frac{5}{33} ) Probability that they have the same color: ( frac{19}{66} ) Probability that at least one is red: ( frac{5}{11} ) Probability that exactly one is yellow: ( frac{16}{33} )

By following these steps, you can apply the principles of combinations and probability in practical scenarios, such as selecting marbles from a bag.