Suppose that the random variable X takes on the values: -1, 0, and 2 with probability $\frac{1}{8}$, $\frac{1}{2}$ and $\frac{3}{8}$. Find the expected value of X.
Understanding Expected Value of a Random Variable
The expected value of a discrete random variable represents the weighted average of all possible values that the random variable can take. The weights used are the probabilities associated with each value. It's often denoted as E[X].
For a discrete random variable X that can take values $x_1, x_2, \dots, x_n$ with corresponding probabilities $P(X=x_1), P(X=x_2), \dots, P(X=x_n)$, the expected value is calculated using the formula:
$ E[X] = \sum_{i=1}^{n} x_i P(X=x_i) $This means we multiply each possible value of the random variable by its probability and then sum up all these products.
Calculating Expected Value for X
In this specific problem, the random variable X takes on the following values with their respective probabilities:
- Value $x_1 = -1$ with Probability $P(X=-1) = \frac{1}{8}$
- Value $x_2 = 0$ with Probability $P(X=0) = \frac{1}{2}$
- Value $x_3 = 2$ with Probability $P(X=2) = \frac{3}{8}$
Step-by-Step Calculation
To find the expected value E[X], we apply the formula:
First, multiply each value by its probability:
- $(-1) \times P(X=-1) = (-1) \times \frac{1}{8} = -\frac{1}{8}$
- $(0) \times P(X=0) = (0) \times \frac{1}{2} = 0$
- $(2) \times P(X=2) = (2) \times \frac{3}{8} = \frac{6}{8}$
Next, sum these products to find the expected value:
$ E[X] = \left(-\frac{1}{8}\right) + (0) + \left(\frac{6}{8}\right) $ $ E[X] = \frac{-1 + 0 + 6}{8} $ $ E[X] = \frac{5}{8} $Therefore, the expected value of the random variable X is $\frac{5}{8}$.
Important Questions from Probability (Notes)
- In a box there are 4 white balls and 6 black balls. A ball is drawn at random. If it is white, it is put back along with two more white balls in the box. If it is black, it is put back in the box and then two black balls are thrown out of the box. Now a ball is drawn again at random from the box. Then, what is the probability that it is black?
- A fair coin is tossed three times. Let A be the event of getting exactly two heads and B be the event of getting at most
two tails, then P(A$\cup$B) is: Bag A contains 3 Red and 4 Black balls while Bag B contains 5 Red and 6 Black balls. One ball is drawn at random from one of the bags and is found to be red. Then, the probability that it was drawn from Bag B is
- If we twice flip a balanced coin, what is the probability of getting at least one head?
1. 1/4
2. 2/4
3. 1/6
4. 3/4 If the probability function for a random variable x is given as f(x) = (x+3)/15 when x = 1, 2 and 3. Find the sum of the values of the probability distribution for x.
