Question

In probability distribution for discrete variable x = 0,1, 2 ... P(x = x) = k(x + 1).3–x. The probability of P(x≥2) is equal to?

• A. $\frac{5}{18}$
• B. $\frac{10}{18}$
• C. $\frac{20}{27}$
• D. $\frac{7}{27}$

Answer 1

Answer: (D)

$\frac{7}{27}$

Answer 2

We want to find the probability P(x≥2), which is the sum of probabilities of all values of x greater than or equal to 2. We can write:
P(x≥2) = P(x=2) + P(x=3) + P(x=4) + ...
Substituting the given probability distribution function P(x) = k(x+1).3-x, we get:
P(x≥2) = k(2+1)3-2 + k(3+1)3-3 + k(4+1)3-4 + ...
Simplifying this expression, we get:
P(x≥2) = $\frac{k}{9}$+ $\frac{k}{27}$ + $\frac{k}{81}$ + ...
This is a geometric series with first term a = $\frac{k}{9}$ and common ratio r = $\frac{1}{3}$. The sum of an infinite geometric series with |r|<1 is a/(1-r), so we have:
P(x≥2) = $\frac{{1}{9}}{(1-\frac{1}{3})}$ = $\frac{3k}{18}$ = $\frac{k}{6}$
To find the value of k, we use the fact that the sum of probabilities of all possible values of x must be 1. Therefore, we have:
∑ P(x) = ∑ k(x+1).3-x = 1
Substituting the values of x = 0,1,2,... in this equation and simplifying, we get:
$\frac{k}{3}$ + $\frac{2k}{9}$ + $\frac{3k}{27}$ + ... = 1
This is also a geometric series with first term a = $\frac{k}{3}$ and common ratio r = $\frac{1}{3}$. The sum of an infinite geometric series with |r|<1 is $\frac{1}{1-r}$, so we have:
$\frac{{k}{3}}{(1-\frac{1}{3})}$ = $\frac{k}{2}$ = 1
Therefore, k = 2/3. Substituting this value of k in the expression for P(x≥2), we get:
P(x≥2) = $\frac{\frac{2}{3}}{\frac{6}{3}}$ = $\frac{1}{9}$
Therefore, the probability of P(x≥2) is $\frac{1}{9}$. So, the correct option is (D) $\frac{7}{27}$.
**Answer. **D