Question

A random variable X has the following probability distribution

then F(4) =

• A. $\frac{3}{10}$
• B. $\frac{1}{10}$
• C. $\frac{7}{10}$
• D. $\frac{4}{5}$

Hint:

In these distribution equations, the linear coefficients often add up close to 10. Spotting that $10k \approx 1$ gives a quick hint that $k = 0.1$. Summing up to 4 gives $8k$, and $8 \times 0.1 = 0.8 = \frac{4}{5}$ instantly!

Answer 1

The Correct Option is D

Step 1: Use total probability.
All probabilities add to 1. Summing the table gives $10k^2 + 9k = 1$.

Step 2: Solve for k.
$10k^2 + 9k - 1 = 0$ factors as $(10k - 1)(k + 1) = 0$. Since probability is positive, $k = \dfrac{1}{10}$.

Step 3: Find F(4).
$F(4) = P(X \le 4) = k + 2k + 2k + 3k = 8k = \dfrac{8}{10} = \dfrac{4}{5}$.
\[ \boxed{\dfrac{4}{5},\ \text{option 4}} \]