Let X be a binomial random variable with parameters $n = 6$ and $p = 0.4$. We have
$P(X=0)=0.0467$
$P(X=1)=0.1866$
$P(X=2)=0.3110$
$P(X=2)=0.2765$
$P(X=2)=0.1382$
$P(X=2)=0.0369$
$P(X=6)=0.0041$
Why is $P(X=6)$ $\neq$ $1-P(X=0)$?
Say we have a sample of $6$ resistors and probability that a resistor functions = $0.4$. Then there is a $P(X=0)=0.0467$ that no resistors are functioning and there is a $P(X=6)=0.0041$ that all resistors are functioning. So why is $P(X=6$) $\neq$ $1-P(X=0)$?
Thank you.
It is possible that only some resistor fail but not all resistor fail. For example, it is possible that exactly $2$ resistor fails.
Also, those values do sum up to $1$.