3.For what value of c, f(x) = c(1⁄4)
x
, x = 0, 1, 2 ... can serve as a probability distribution of X?

Hence find (a) P(X < 2) (b) P(X > 1) (c) P(3 ≤ X < 5)

### Problem: 
We are given a function \( f(x) = c \left( \frac{1}{4} \right)^x \), where \( x = 0, 1, 2, \dots \). We are asked to find the value of \( c \) such that this function serves as a probability distribution of \( X \), and to calculate some probabilities based on that distribution.

### Step 1: Find the value of \( c \)
For \( f(x) \) to be a probability distribution, the sum of probabilities over all possible values of \( X \) must be equal to 1. That is,

\[
\sum_{x=0}^{\infty} f(x) = 1.
\]

Substitute \( f(x) = c \left( \frac{1}{4} \right)^x \):

\[
\sum_{x=0}^{\infty} c \left( \frac{1}{4} \right)^x = 1.
\]

This is an infinite geometric series with the first term \( a = c \) and the common ratio \( r = \frac{1}{4} \). The sum of an infinite geometric series is given by:

\[
S = \frac{a}{1 - r}.
\]

Using this formula, we get:

\[
\sum_{x=0}^{\infty} c \left( \frac{1}{4} \right)^x = \frac{c}{1 - \frac{1}{4}} = \frac{c}{\frac{3}{4}} = \frac{4c}{3}.
\]

Set this equal to 1:

\[
\frac{4c}{3} = 1 \quad \Rightarrow \quad c = \frac{3}{4}.
\]

Thus, the value of \( c \) is \( \frac{3}{4} \).

### Step 2: Find the required probabilities

Now that we have the probability distribution \( f(x) = \frac{3}{4} \left( \frac{1}{4} \right)^x \), we can calculate the required probabilities.

#### (a) \( P(X < 2) \)
We need to find \( P(X = 0) + P(X = 1) \):

\[
P(X = 0) = \frac{3}{4} \left( \frac{1}{4} \right)^0 = \frac{3}{4},
\]
\[
P(X = 1) = \frac{3}{4} \left( \frac{1}{4} \right)^1 = \frac{3}{16}.
\]

Thus,

\[
P(X < 2) = P(X = 0) + P(X = 1) = \frac{3}{4} + \frac{3}{16} = \frac{12}{16} + \frac{3}{16} = \frac{15}{16}.
\]

#### (b) \( P(X > 1) \)
We need to find the sum for \( x = 2, 3, 4, \dots \). Using the complement rule:

\[
P(X > 1) = 1 - P(X \leq 1) = 1 - \left( P(X = 0) + P(X = 1) \right).
\]

We already know that \( P(X = 0) + P(X = 1) = \frac{15}{16} \), so

\[
P(X > 1) = 1 - \frac{15}{16} = \frac{1}{16}.
\]

#### (c) \( P(3 \leq X < 5) \)
We need to find \( P(X = 3) + P(X = 4) \):

\[
P(X = 3) = \frac{3}{4} \left( \frac{1}{4} \right)^3 = \frac{3}{4} \times \frac{1}{64} = \frac{3}{256},
\]
\[
P(X = 4) = \frac{3}{4} \left( \frac{1}{4} \right)^4 = \frac{3}{4} \times \frac{1}{256} = \frac{3}{1024}.
\]

Thus,

\[
P(3 \leq X < 5) = P(X = 3) + P(X = 4) = \frac{3}{256} + \frac{3}{1024} = \frac{12}{1024} + \frac{3}{1024} = \frac{15}{1024}.
\]

### Summary of Results:
- \( c = \frac{3}{4} \)
- \( P(X < 2) = \frac{15}{16} \)
- \( P(X > 1) = \frac{1}{16} \)
- \( P(3 \leq X < 5) = \frac{15}{1024} \)

Would you like more details or have any questions?

#### Related Questions:
1. How do you derive the sum of an infinite geometric series?
2. Can you explain the complement rule in probability?
3. What is the difference between discrete and continuous probability distributions?
4. How can we use probability distributions to model real-world scenarios?
5. How do you generalize the sum of probabilities for \( P(a \leq X \leq b) \) for larger ranges?

#### Tip:
Always check that the sum of all probabilities in a distribution adds up to 1 to ensure it's valid.

For what value of c, f(x) = c(1/4)^x, x = 0, 1, 2 ... can serve as a probability distribution of X? Hence find (a) P(X < 2) (b) P(X > 1) (c) P(3 ≤ X < 5)

Find the value of c for which f(x) = c(1/4)^x serves as a probability distribution. Learn how to calculate key probabilities such as P(X < 2), P(X > 1), and P(3 ≤ X < 5) using geometric series and probability theorems. Suitable for undergraduate students in probability courses.

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