Question

The probability mass function of a random variable $X$ is given by: $$ P(X = r) = \begin{cases} \frac{{}^nC_r}{32}, & r = 0,1,2,\dots,n \\ 0, & \text{otherwise} \end{cases} $$ Find: $$ P(X \le 2) $$ 

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Hint:

Logic Tip: A symmetric binomial distribution (like flipping 5 fair coins) has probabilities that mirror each other: $P(X=0)=P(X=5)$, $P(X=1)=P(X=4)$, etc. Since $P(X \le 2)$ covers exactly half the symmetrical distribution values relative to their weights, it must equal $0.5$.

Answer 1

The Correct Option is B

Concept:
For a valid probability mass function (PMF), the sum of all probabilities over the entire sample space must equal 1. $$\sum_{r=0}^{n} P(X=r) = 1$$ We also use the binomial theorem property that the sum of all binomial coefficients for a given $n$ is $\sum_{r=0}^{n} {}^nC_r = 2^n$.
Step 1: Find the value of n.
Summing the probabilities for all possible values of $r$ from $0$ to $n$: $$\sum_{r=0}^{n} \frac{^nC_r}{32} = 1$$ Factor out the constant denominator: $$\frac{1}{32} \sum_{r=0}^{n} {}^nC_r = 1$$ Substitute the binomial property $\sum_{r=0}^{n} {}^nC_r = 2^n$: $$\frac{2^n}{32} = 1$$ $$2^n = 32$$ Since $32 = 2^5$, we find that $n = 5$.
Step 2: Calculate the required probability $P(X \le 2)$.
The question asks for $P(X \le 2)$, which is the sum of probabilities for $r = 0, 1, 2$: $$P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$$ Substitute the PMF formula with $n=5$: $$P(X \le 2) = \frac{^5C_0}{32} + \frac{^5C_1}{32} + \frac{^5C_2}{32}$$ Evaluate the combinations: $${}^5C_0 = 1$$ $${}^5C_1 = 5$$ $${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$$ Add the terms: $$P(X \le 2) = \frac{1 + 5 + 10}{32} = \frac{16}{32}$$ Simplify the fraction: $$P(X \le 2) = \frac{1}{2}$$