If the independent random variables X,Y are Binomially distributed with n = 3, p = \(\dfrac{1}{3}\) and n = 5, p = \(\dfrac{1}{3}\) respectively, then the probability of (X + Y ≥ 1) is:
1 - \((\dfrac{2}{3})^8\)
Step 1 — Use the additive property of independent Binomials with equal p:
If \(X \sim B(3, 1/3)\) and \(Y \sim B(5, 1/3)\) are independent, then \(Z = X+Y \sim B(8, 1/3)\).
Step 2 — Apply the complement rule:
\[P(Z \ge 1) = 1 - P(Z = 0)\]
Step 3 — Compute P(Z = 0):
\[P(Z=0) = \binom{8}{0}\left(\dfrac{1}{3}\right)^0\left(\dfrac{2}{3}\right)^8 = \left(\dfrac{2}{3}\right)^8\]
Therefore \(P(X+Y \ge 1) = 1 - \left(\dfrac{2}{3}\right)^8\).
Similar Questions
Let {X i, i ≥ 1} be independent and identically distributed random variables with (X i= 1) = p = 1 - P(X i= 0), S n= X i. The distribution of S nis?
If X has Binomial distribution with parameters n and p such that np =λ, then \(\mathop {\lim }\limits_{n \to \infty } b\left( {x,n,p} \right);x = 0,1,2,.....\) is equal to:
A random variable X is binomially distributed with parameter n = 25 and p = 0.2. The skewness of the variable X is
Important Questions from Binomial Distribution
In a Binomial distribution, the mean is three times its variance. What is the probability of exactly 3 successes out of 5 trials?
If mean and variance of a Binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is
For Binomial distribution, n = 10 and p = 0.6, E(X 2) (second moment about origin) is:
Find out the fallacy if any in the statement:
“The mean and the variance of a binomial distribution is 16.2 and 29.4 respectively.”
If log e(x) is normally distributed with mean 1 and variance 4, then P(0.5 < x < 2) is:
(where the area between z = 0 and z = 0.25 is 0.0987 and the area between z = 0 and z = 0.5 is 0.1915)
