Every light bulb is defective with 2% probability. What is the probability that a package of 8 bulbs will not suffice for a project requiring 7? I am trying to solve this problem on Bernoulli distribution but looks like there is some gap my understanding.
To me this equivalent to finding out $1.0 - [P(A) + P(B)]$,
where event A = package having no defective light bulb and event B = package having one defective light bulb
$P(A) = (0.98)^8$
$P(B) = (0.98)^7 \cdot 0.02$
P(A) + P(B) = $(0.98)^7$
probability that a package of 8 bulbs will not suffice for a project requiring 7 $= 1.0 - (0.98)^7 = 0.13187$
Auto-grader indicates the answer is incorrect.
 A: Your strategy is correct.  However, you did not account for which of the eight bulbs is defective when exactly one of the eight bulbs is defective.
The binomial distribution states that the probability of exactly $k$ successes in $n$ trials each with probability $p$ of success is
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ is the number of ways exactly $k$ successes could occur in $n$ trials.
If we let $X = k$ denote the event that exactly $k$ of the light bulbs work properly, then the probability that eight light bulbs will not suffice is the probability that fewer than seven of the bulbs work properly (or, equivalently, that at least two are defective), which is $\Pr(X < 7)$ where $p = 0.98$ and $n = 8$.
\begin{align*}
\Pr(X < 7) & = 1 - \Pr(X \geq 7)\\
           & = 1 - \Pr(X = 7) - \Pr(x = 8)\\
           & = 1 - \binom{8}{7}(0.98)^7(0.02)^1 - \binom{8}{8}(0.98)^8(0.02)^0\\
              & \approx 0.0103368921
\end{align*}
