Addition Rule for Mutually Exclusive Events

Probability of selecting a defective bulb ($$p$$) = $$\frac{10}{100} = 0.1$$

Probability of selecting a non-defective bulb ($$q$$) = $$1 - p = 0.9$$

Number of trials ($$n$$) = 8 (Because 8 bulbs are selected)

Let $$X$$ be the random variable representing the number of defective bulbs. 

$$X$$ follows a binomial distribution: $$X \sim B(n, p)$$.

$$P(X = k) = \binom{n}{k} p^k q^{n-k}$$

We need to find $$P(X \ge 7)$$, which is the sum of the probabilities of getting exactly 7 or exactly 8 defective bulbs:

$$P(X \ge 7) = P(X = 7) + P(X = 8)$$

For $$k = 7$$: $$P(X = 7) = \binom{8}{7} (0.1)^7 (0.9)^{8-7}$$

$$P(X = 7) = 8 \times (0.1)^7 \times 0.9$$

$$P(X = 7) = 7.2 \times (0.1)^7$$

$$P(X = 7) = 0.00000072$$

For $$k = 8$$: $$P(X = 8) = \binom{8}{8} (0.1)^8 (0.9)^{8-8}$$

$$P(X = 8) = 1 \times (0.1)^8 \times 1$$

$$P(X = 8) = 0.1 \times (0.1)^7$$

$$P(X = 8) = 0.00000001$$

Final Sum: $$P(X \ge 7) = 7.2 \times 10^{-7} + 0.1 \times 10^{-7}$$

$$P(X \ge 7) = 7.3 \times 10^{-7}$$

The probability of getting at least 7 defective bulbs is $$7.3 \times 10^{-7}$$ (or $$\frac{73}{10^8}$$).

The following are the cases possible when the dies are thrown and the sum of the numbers on the two faces is $$4$$:

1 and 3
First die 1 and the second die 3 = $$\dfrac{2}{6}\times \dfrac{2}{6} = \dfrac{4}{26}$$
First die 3 and the second die 1 = $$\dfrac{1}{6}\times \dfrac{1}{6} = \dfrac{1}{36}$$

2 and 2
First die 2 and the second die 2 = $$\dfrac{2}{6}\times \dfrac{2}{6} = \dfrac{4}{36}$$

The probability of getting a sum of $$4$$ in all: $$\dfrac{4+1+4}{36} = \dfrac{1}{4}$$

The following are the cases possible when the dies are thrown and the sum of the numbers on the two faces is $$5$$:

1 and 4
First die 1 and the second die 4: $$\dfrac{2}{6}\times \dfrac{1}{6} = \dfrac{2}{36}$$
First die 4 and the second die 1: $$\dfrac{1}{6}\times \dfrac{1}{6} = \dfrac{1}{36}$$

2 and 3
First die 2 and the second die 3: $$\dfrac{2}{6}\times \dfrac{2}{6} = \dfrac{4}{36}$$
First die 3 and the second die 2: $$\dfrac{1}{6}\times \dfrac{2}{6} = \dfrac{2}{36}$$

The probability of getting a sum of $$5$$ in all: $$\dfrac{2+1+4+2}{36} = \dfrac{1}{4}$$

Thus, the combined probability of getting a sum of either $$4$$ or $$5$$ when the two dice are thrown is $$\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}$$