A bag contains 4 yellow, 2 red, and 6 green marbles. Two marbles are drawn. The first is replaced before the second is drawn. A random variable assigns the number of red marbles to each outcome. Calculate the expected value of the random variable
P(red) = 2/(4+2+6) = ?
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
P(red) * P(red) = ?
0.67
To calculate the expected value of the random variable, we need to find the probabilities of all possible outcomes and multiply them by their corresponding values.
Let's break down the problem step by step:
Step 1: Find the probabilities of each outcome.
Since the first marble is replaced before the second is drawn, the probability of drawing a red marble at any draw is the same.
The probability of drawing a red marble in a single draw can be calculated as:
P(Red) = (number of red marbles) / (total number of marbles)
= 2 / (4 + 2 + 6)
= 2 / 12
= 1 / 6
So, the probability of drawing a red marble in a single draw is 1/6.
Step 2: Calculate the expected value.
The expected value (E) of a random variable is calculated by summing the products of each outcome and its probability.
Let's calculate the expected value (E) for the random variable, where X represents the number of red marbles:
E(X) = (probability of getting X = 0) * (value of X = 0) +
(probability of getting X = 1) * (value of X = 1) +
(probability of getting X = 2) * (value of X = 2)
Here, the values of X are 0, 1, and 2, which represent the number of red marbles drawn.
E(X) = (P(X = 0) * 0) + (P(X = 1) * 1) + (P(X = 2) * 2)
Step 3: Calculate the probabilities of each outcome:
P(X = 0): This means drawing 0 red marbles.
Since the first marble is replaced before the second is drawn, the probability of not drawing any red marble in both draws is calculated as:
P(X = 0) = (probability of not drawing a red marble in a single draw) * (probability of not drawing a red marble in the second draw)
= P(Red)' * P(Red)'
= (1 - P(Red))^2
= (1 - 1/6)^2
= (5/6)^2
= 25/36
P(X = 1): This means drawing 1 red marble.
Since the first marble is replaced before the second is drawn, the probability of drawing one red marble in one of the draws and not drawing a red marble in the other draw is calculated as:
P(X = 1) = (probability of drawing a red marble in the first draw) * (probability of not drawing a red marble in the second draw)
= P(Red) * P(Red') + P(Red') * P(Red)
= P(Red) * P(Red') + P(Red') * P(Red)
= P(Red) * P(Red') + P(Red') * P(Red)
= 1/6 * 5/6 + 5/6 * 1/6
= 10/36
P(X = 2): This means drawing 2 red marbles.
Since the first marble is replaced before the second is drawn, the probability of drawing a red marble in both draws is calculated as:
P(X = 2) = (probability of drawing a red marble in a single draw) * (probability of drawing a red marble in the second draw)
= P(Red) * P(Red)
= (1/6)^2
= 1/36
Step 4: Substitute the probabilities into the expected value equation:
E(X) = (P(X = 0) * 0) + (P(X = 1) * 1) + (P(X = 2) * 2)
= (25/36 * 0) + (10/36 * 1) + (1/36 * 2)
= 0 + 10/36 + 2/36
= 12/36
= 1/3
Therefore, the expected value of the random variable is 1/3.