If the probability function for a random variable \( x \) is given as \( f(x) = \frac{x+3}{15} \) when \( x = 1, 2, 3 \), find the sum of the values of the probability distribution for \( x \).
Step 1: Understand the sum of probabilities.
The sum of the probabilities for a discrete random variable must always be equal to 1. The probability mass function is given as:
\[
f(x) = \frac{x+3}{15} \text{for } x = 1, 2, 3.
\]
We need to check if the sum of probabilities equals 1.
Step 2: Calculate the probabilities.
- For \( x = 1 \), \( f(1) = \frac{1+3}{15} = \frac{4}{15} \)
- For \( x = 2 \), \( f(2) = \frac{2+3}{15} = \frac{5}{15} \)
- For \( x = 3 \), \( f(3) = \frac{3+3}{15} = \frac{6}{15} \)
Step 3: Check the sum.
The sum of the probabilities is:
\[
\frac{4}{15} + \frac{5}{15} + \frac{6}{15} = \frac{15}{15} = 1.
\]
Step 4: Conclusion.
The sum of the probabilities is 1, so the correct answer is (D).
A weight of $500\,$N is held on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P$ acting at $30^\circ$ to the inclined plane as shown. Then the value of force $P$ is:
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