A random variable X takes the values 1, 2, 3 and 4 such that

P(X=1)=P(X=2)=2P(X=3)=3P(X=4).Write the probability distribution of X and find

(i) P(X>2), (ii) P(1<X<4//X>2), (iii)P(X= 1or 2).

To solve this, let's first express the probability distribution of the random variable \( X \).

### Step 1: Express the probabilities in terms of a common variable

Let the probability \( P(X = 1) = P(X = 2) = p \).

From the given conditions:
\[
P(X = 3) = \frac{p}{2}, \quad P(X = 4) = \frac{p}{3}.
\]

Now, the total probability must sum to 1:
\[
P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1.
\]
Substitute the probabilities:
\[
p + p + \frac{p}{2} + \frac{p}{3} = 1.
\]
Simplify the left-hand side:
\[
2p + \frac{p}{2} + \frac{p}{3} = 1.
\]
To combine these, find the least common denominator, which is 6:
\[
2p = \frac{12p}{6}, \quad \frac{p}{2} = \frac{3p}{6}, \quad \frac{p}{3} = \frac{2p}{6}.
\]
Now add the fractions:
\[
\frac{12p + 3p + 2p}{6} = 1 \quad \Rightarrow \quad \frac{17p}{6} = 1.
\]
Solve for \( p \):
\[
p = \frac{6}{17}.
\]

Thus, the probabilities are:
\[
P(X = 1) = P(X = 2) = \frac{6}{17}, \quad P(X = 3) = \frac{3}{17}, \quad P(X = 4) = \frac{2}{17}.
\]

### Step 2: Find the required probabilities

#### (i) \( P(X > 2) \)
This is the probability that \( X = 3 \) or \( X = 4 \):
\[
P(X > 2) = P(X = 3) + P(X = 4) = \frac{3}{17} + \frac{2}{17} = \frac{5}{17}.
\]

#### (ii) \( P(1 < X < 4 \mid X > 2) \)
We are asked to find the conditional probability \( P(1 < X < 4 \mid X > 2) \).

First, find \( P(1 < X < 4) \), which is the probability that \( X = 3 \):
\[
P(1 < X < 4) = P(X = 3) = \frac{3}{17}.
\]
Now, the probability \( P(X > 2) \) is already computed as \( \frac{5}{17} \).

So the conditional probability is:
\[
P(1 < X < 4 \mid X > 2) = \frac{P(1 < X < 4 \cap X > 2)}{P(X > 2)} = \frac{P(X = 3)}{P(X > 2)} = \frac{\frac{3}{17}}{\frac{5}{17}} = \frac{3}{5}.
\]

#### (iii) \( P(X = 1 \text{ or } 2) \)
This is the probability that \( X = 1 \) or \( X = 2 \):
\[
P(X = 1 \text{ or } 2) = P(X = 1) + P(X = 2) = \frac{6}{17} + \frac{6}{17} = \frac{12}{17}.
\]

### Final Answers:
1. \( P(X > 2) = \frac{5}{17} \)
2. \( P(1 < X < 4 \mid X > 2) = \frac{3}{5} \)
3. \( P(X = 1 \text{ or } 2) = \frac{12}{17} \)

---

Would you like further details on any part of the solution?

Here are some related questions you might find interesting:
1. How do you compute the expected value of \( X \)?
2. What is the variance of \( X \)?
3. How do conditional probabilities work in different types of probability distributions?
4. What is the cumulative distribution function (CDF) for \( X \)?
5. How would you solve if the probabilities were not equal at first?

**Tip**: Conditional probabilities are always calculated by dividing the joint probability by the probability of the given condition.

A random variable X takes the values 1, 2, 3, and 4 such that P(X=1)=P(X=2)=2P(X=3)=3P(X=4). Write the probability distribution of X and find (i) P(X>2), (ii) P(1<X<4//X>2), (iii) P(X=1 or 2).

Learn how to determine the probability distribution of a random variable X with values 1, 2, 3, and 4, where P(X=1)=P(X=2)=2P(X=3)=3P(X=4). Find P(X>2), P(1<X<4//X>2), and P(X=1 or 2). This solution covers core probability concepts, including conditional probability and total probability, suitable for Grades 10-12.

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