Understanding the Probability of Drawing Marbles in Any Order
In this article, we will delve into the probability of drawing two red marbles and one blue marble in any order from a box containing 6 red marbles and 4 blue marbles. We'll explore this through different methods, including basic probability, combinations, and the binomial distribution. Let's begin.Introduction to the Problem
A box contains 6 red marbles and 4 blue marbles. One marble is randomly drawn, its color is noted, and the marble is put back into the box. This procedure is repeated two more times. We want to find the probability of getting two red marbles and one blue marble in any order.Method 1: Basic Probability
We can first calculate the probability of drawing a red marble and a blue marble in each step and then combine these probabilities to find the desired result. Calculate the probability of each draw:The probability of drawing a red marble (R) is:
[ PR frac{6}{10} 0.6 ]The probability of drawing a blue marble (B) is:
[ PB frac{4}{10} 0.4 ] Identify the different orders: The possible arrangements are RRB, RBR, and BRR. These are the permutations of drawing two red marbles and one blue marble in any order. Calculate the probability of each arrangement: The probability of drawing RRB is: [ PRRB PR times PR times PB 0.6 times 0.6 times 0.4 0.144 ] The probabilities for the other arrangements RBR and BRR are the same, so we have: [ PRBR PBRR 0.144 ] Combine the probabilities: To find the total probability of getting two red marbles and one blue marble in any order, we add the probabilities of all three arrangements: [ P(2 , Red , 1 , Blue) PRRB PRBR PBRR 0.144 0.144 0.144 0.432 ] Final answer: The probability of drawing two red marbles and one blue marble in any order is boxed{0.432}.