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.
This question asks us to calculate the sum of the probabilities for a given probability function of a discrete random variable. The random variable $x$ can take values 1, 2, and 3, and its probability function is defined as $f(x) = (x+3)/15$.
Probability Function Explained
A probability function, often denoted as $f(x)$ or $P(x)$, assigns a probability to each possible value that a random variable can take. For a valid probability distribution, two main conditions must be met:
- The probability for each value must be non-negative: $f(x) ≥ 0$ for all $x$.
- The sum of probabilities for all possible values of the random variable must equal 1: $Σ f(x) = 1$.
Calculating Probabilities for Each Value of x
We need to calculate the probability for each specific value that the random variable $x$ can assume (1, 2, and 3) using the given function $f(x) = (x+3)/15$.
- For x = 1: Substitute $x=1$ into the function: $f(1) = (1 + 3) / 15 = 4 / 15$
- For x = 2: Substitute $x=2$ into the function: $f(2) = (2 + 3) / 15 = 5 / 15$
- For x = 3: Substitute $x=3$ into the function: $f(3) = (3 + 3) / 15 = 6 / 15$
Probability Distribution Table
We can summarize these probabilities in a table:
| Value of x | Probability f(x) |
|---|---|
| 1 | $4/15$ |
| 2 | $5/15$ |
| 3 | $6/15$ |
Summing the Probabilities
To find the sum of the values of the probability distribution, we add the probabilities calculated for each possible value of $x$:
Sum = $f(1) + f(2) + f(3)$
Sum = $(4/15) + (5/15) + (6/15)$
Sum = $(4 + 5 + 6) / 15$
Sum = $15 / 15$
Sum = $1$
The sum of the probabilities for all possible values of the random variable $x$ is 1, which confirms that this is a valid probability distribution.
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 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.
