Q.The number of defective bulbs $X$ found when 3 bulbs are randomly checked from a large batch has the probability distribution: $P(X=0)=0.512$, $P(X=1)=0.384$, $P(X=2)=0.096$, $P(X=3)=0.008$.

After confirming the probabilities sum to 1, the c.d.f. is built by cumulative addition, and the required probabilities are read directly from it (with $P(X\geq1)$ found using the complement of $P(X=0)$). …

1. Verifying validity: All four probabilities are non-negative, and

   $0.512+0.384+0.096+0.008 = 1.000$

   so this is a valid probability distribution.
2. Constructing the c.d.f.:

   $F(0) = 0.512$

   $F(1) = 0.512+0.384 = 0.896$

   $F(2) = 0.896+0.096 = 0.992$

   $F(3) = 0.992+0.008 = 1.000$

As required, $F(x)$ rises steadily and reaches exactly $1$ at the largest value.

(iii) Finding the required probabilities:

For $P(X\geq1)$, it is quickest to use the complement rule, since "$X\geq1$" is everything except "$X=0$":

$P(X\geq1) = 1 - P(X=0) = 1-0.512 = 0.488$ …

$P(X\geq1)$ can also be found by direct addition of $P(X=1)+P(X=2)+P(X=3)$ instead of the complement rule, as shown in the cross-check — the complement rule is simply the quicke …

A common mistake is computing $P(X\geq1)$ as $1-F(1)$ instead of $1-P(X=0)$ — since $F(1)$ already includes $P(X=0)$ and $P(X=1)$ together, subtracting it from 1 wrongly excludes $X=1$ from the 'at least on …