Animated Solution for Mathematics - Probability: The probability distribution of a random variable $X$ is given below : $$\begin{array}{ccccccccc}X& 4k& \frac{30}{7}k& \frac{32}{7}k& \frac{34}{7}k& \frac{36}{7}k& \frac{38}{7}k& \frac{40}{7}k& 6k\\ P(X)& \frac{2}{15}& \frac{1}{15}& \frac{2}{15}& \frac{1}{5}& \frac{1}{15}& \frac{2}{15}& \frac{1}{5}& \frac{1}{15}\end{array}$$ If $E(X)=\frac{263}{15}$, then $P(X<20)$ is equal to :

Welcome, fellow explorer of probability! Today, we are going to dissect a problem that might look like a tedious table of numbers, but is actually a beautiful exercise in algebraic elegance.

We are given a probability distribution for a random variable $X$, where the values of $X$ are scaled by an unknown constant $k$. Our mission is to find $P(X<20)$, but to get there, we must first uncover the identity of $k$ using the given expected value $E(X)=\frac{263}{15}$.

The Weighted Average

What is the expected value? Think of it as the 'center of mass' of your probability distribution. It is the weighted average of all possible values of $X$, where each value is weighted by its probability.

Mathematically, we define it as:

Imagine you are standing on a balance beam. The values of $X$ are positions on the beam, and the probabilities $P(X)$ are the masses placed at those positions. The expected value is the point where the beam would perfectly balance.

Our table gives us eight such points. To find $E(X)$, we must multiply each $X_i$ by its corresponding $P(X_i)$ and sum them all up. It sounds like a mountain of arithmetic, but let's see how we can climb it efficiently.

The Algebraic Dance

Let's write out the sum: $E(X)=(4k\cdot \frac{2}{15})+(\frac{30}{7}k\cdot \frac{1}{15})+(\frac{32}{7}k\cdot \frac{2}{15})+(\frac{34}{7}k\cdot \frac{1}{5})+\dots$ and so on.

Stop! Do you see the pattern? Every term has a $k$ in the numerator and a $15$ in the denominator. Even the terms with $\frac{1}{5}$ can be written as $\frac{3}{15}$.

This is our golden opportunity. Let's factor out $\frac{k}{15}$ from the entire expression:

By grouping the terms with a denominator of $7$ and adding the integers $8$ and $6$ separately, the expression simplifies beautifully. The sum of the numerators over $7$ is $\frac{428}{7}$, and adding the integers $14$ gives us $\frac{98+428}{7}=\frac{526}{7}$.

Thus, our expected value is:

The Reveal of $k$

Now, we equate this to the given value:

Look at the numbers. They aren't random; they are designed to cancel. Notice that $526=263\times 2$ and $105=15\times 7$.

When we solve for $k$, we get:

We have found our key!

The Final Condition

With $k=\frac{7}{2}$, we can now find the actual values of $X$. Substituting $k$ back into the table, we get the sequence: $14,15,16,17,18,19,20,21$.

Our objective is $P(X<20)$. Looking at our sequence, the values that satisfy this are $14,15,16,17,18,$ and $19$.

We simply sum the probabilities corresponding to these values:

And there it is! The probability is $\frac{11}{15}$. A journey through algebra that ends in a clean, satisfying result. Keep practicing, and remember: the complexity is just a mask for the underlying simplicity.