Probability of Drawing One Red and One White Ball Without Replacement
In probability theory, understanding how to calculate the probability of specific events is a fundamental concept. This article explores the calculation of the probability of drawing one red and one white ball from a bag without replacement, providing a step-by-step explanation and key formulas.
Introduction to Probability Basics
Probability is a measure of the likelihood that an event will occur. It is defined as the number of favorable outcomes divided by the total number of possible outcomes. When drawing balls from a bag without replacement, the probability of subsequent draws changes due to the reduction in the total number of balls.
Probability Calculation Steps
1. Initial Setup
Consider a bag containing 4 white and 6 red balls, making a total of 10 balls. We aim to find the probability of drawing one red ball and one white ball without replacement.
2. Total Ways to Draw Two Balls
The total number of ways to draw 2 balls from 10 can be calculated using combinations. The formula for combinations is:
[ binom{n}{k} frac{n!}{k!(n-k)!} ]For our scenario:
[ binom{10}{2} frac{10 times 9}{2 times 1} 45 ]3. Ways to Draw One Red and One White
We need to account for the different orders in which we can draw the balls.
Scenario 1: Red First, Then White
The number of ways to choose 1 red ball from 6 is:
[ binom{6}{1} 6 ]The number of ways to choose 1 white ball from 4 is:
[ binom{4}{1} 4 ]The total number of ways for this scenario is:
[ 6 times 4 24 ]Scenario 2: White First, Then Red
The number of ways to choose 1 white ball from 4 is:
[ binom{4}{1} 4 ]The number of ways to choose 1 red ball from 6 is:
[ binom{6}{1} 6 ]The total number of ways for this scenario is:
[ 4 times 6 24 ]4. Total Ways to Get One Red and One White
Since these are the only two scenarios that give us one red and one white ball, we sum up the ways from both scenarios:
[ 24 24 48 ]5. Probability Calculation
The probability of drawing one red and one white ball is the number of favorable outcomes (48) divided by the total number of ways to draw 2 balls (45):
[ P(text{one red and one white}) frac{48}{45} ]This simplifies to:
[ P(text{one red and one white}) frac{8}{15} ]Conclusion
To summarize, the probability of drawing one red and one white ball from a bag containing 4 white and 6 red balls (10 balls in total) without replacement is ( frac{8}{15} ). This result accounts for the sequential drawing of balls and ensures we do not double count any specific arrangements.