Probability:Binomial Distribution Mean and Variance Problem

Question

Question:

X follows Binomial Distribution with mean $3$ and Variance $\frac32$,find $P(X\ge 1)$.

My approach:

I know that $np=$ mean $=3$ and $npq=$variance$ =3/2$,where $n$ is the number of independent trials in Bernoulli's trials and $p$ refers to the number of success and $q$ refers to the number of failures.I cannot understand the meaning of $P(X\ge 1)$ which prevents me from solving the question further.

A solution with explanation would be very helpful in strengthening my basic concepts.

Answer 1

$X \geq 1$ means $X$ is greater than or equal to $1$.

$P(X \ge 1)$ means the probability that $X$ is greater than or equal to $1$.

Guide:

• Notice that Binomial distribution takes non-negative integer values, hence a quick way to compute this would be to compute $P(X \geq 1 ) = 1-P(X=0)$.

• You might like to use the value of $np$ and $npq=np(1-p)$ to solve for the values of $n$ and $p$ first, from there you can compute $P(X=0)$.

Correction to concept:

$p$ is the probability of success, rather than number of success.