A fair coin is tossed fourvtimes, let x be a random variable representing the number of heads. Find  the P.d.f and c.d.f of x

Given that a fair coin is tossed four times, we can determine the probability distribution function (P.d.f) and cumulative distribution function (C.d.f) of the random variable \( X \), which represents the number of heads.

### 1. Probability Distribution Function (P.d.f)

The number of heads, \( X \), can take values from 0 to 4. The probability of getting \( k \) heads in 4 tosses can be determined using the binomial distribution:

\[
P(X = k) = \binom{4}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{4-k} = \binom{4}{k} \left(\frac{1}{2}\right)^4
\]

Where \( \binom{4}{k} \) is the binomial coefficient.

Let's calculate the probabilities for each possible value of \( X \):

- \( P(X = 0) = \binom{4}{0} \left(\frac{1}{2}\right)^4 = \frac{1}{16} \)
- \( P(X = 1) = \binom{4}{1} \left(\frac{1}{2}\right)^4 = \frac{4}{16} = \frac{1}{4} \)
- \( P(X = 2) = \binom{4}{2} \left(\frac{1}{2}\right)^4 = \frac{6}{16} = \frac{3}{8} \)
- \( P(X = 3) = \binom{4}{3} \left(\frac{1}{2}\right)^4 = \frac{4}{16} = \frac{1}{4} \)
- \( P(X = 4) = \binom{4}{4} \left(\frac{1}{2}\right)^4 = \frac{1}{16} \)

So the P.d.f of \( X \) is:

\[
P(X = k) = 
\begin{cases} 
\frac{1}{16} & \text{if } k = 0 \\
\frac{1}{4} & \text{if } k = 1 \\
\frac{3}{8} & \text{if } k = 2 \\
\frac{1}{4} & \text{if } k = 3 \\
\frac{1}{16} & \text{if } k = 4 \\
0 & \text{otherwise}
\end{cases}
\]

### 2. Cumulative Distribution Function (C.d.f)

The C.d.f \( F_X(k) \) is defined as the probability that \( X \) takes a value less than or equal to \( k \). This is computed by summing the P.d.f up to \( k \).

\[
F_X(k) = P(X \leq k) = \sum_{i=0}^{k} P(X = i)
\]

Calculating \( F_X(k) \) for each \( k \):

- \( F_X(0) = P(X \leq 0) = P(X = 0) = \frac{1}{16} \)
- \( F_X(1) = P(X \leq 1) = P(X = 0) + P(X = 1) = \frac{1}{16} + \frac{4}{16} = \frac{5}{16} \)
- \( F_X(2) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = \frac{1}{16} + \frac{4}{16} + \frac{6}{16} = \frac{11}{16} \)
- \( F_X(3) = P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = \frac{1}{16} + \frac{4}{16} + \frac{6}{16} + \frac{4}{16} = \frac{15}{16} \)
- \( F_X(4) = P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = \frac{1}{16} + \frac{4}{16} + \frac{6}{16} + \frac{4}{16} + \frac{1}{16} = 1 \)

So the C.d.f of \( X \) is:

\[
F_X(k) = 
\begin{cases} 
0 & \text{if } k < 0 \\
\frac{1}{16} & \text{if } 0 \leq k < 1 \\
\frac{5}{16} & \text{if } 1 \leq k < 2 \\
\frac{11}{16} & \text{if } 2 \leq k < 3 \\
\frac{15}{16} & \text{if } 3 \leq k < 4 \\
1 & \text{if } k \geq 4
\end{cases}
\]

Would you like further details or have any questions?

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### Related Questions:
1. How do you derive the binomial coefficient \(\binom{4}{k}\)?
2. What is the expected value of \( X \) for this binomial distribution?
3. How does the variance of \( X \) compare to its expected value?
4. How would the P.d.f and C.d.f change if the coin was biased?
5. What is the probability that the number of heads is exactly 2?
6. What is the probability of getting at least 3 heads?
7. How does the P.d.f change if the number of tosses is increased to 6?
8. How can you verify these results using a simulation?

### Tip:
The binomial distribution is a discrete distribution that arises from the number of successes in a fixed number of independent Bernoulli trials. Understanding the binomial theorem can greatly help in calculating probabilities for such distributions.

Understand the probability distribution function (P.d.f) and cumulative distribution function (C.d.f) of a random variable representing the number of heads in four fair coin tosses. Includes detailed calculations and formulas for the binomial distribution.

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