Problem Overview: Defective Bulb Probability
This problem requires calculating probability using the binomial distribution framework. We are selecting bulbs one by one with replacement, ensuring each selection is an independent event with a constant probability of success (selecting a defective bulb).
- Total bulbs in the lot: 100
- Number of defective bulbs: 10
- Number of non-defective bulbs: 90
- Number of bulbs selected (trials), denoted as n: 8
The probability of success (selecting a defective bulb) in a single trial is:
p = $\frac{\text{Number of defective bulbs}}{\text{Total bulbs}} = \frac{10}{100} = 0.1$
The probability of failure (selecting a non-defective bulb) in a single trial is:
q = $1 - p = 1 - 0.1 = 0.9$
The question asks for the probability of getting at least 7 defective bulbs. This means we need to find the probability of getting exactly 7 defective bulbs OR exactly 8 defective bulbs. Let $X$ be the number of defective bulbs selected. We need to calculate $P(X \ge 7)$.
$P(X \ge 7) = P(X=7) + P(X=8)$
Binomial Probability Calculation
The formula for binomial probability is $P(X=k) = C(n, k) \times p^k \times q^{(n-k)}$, where $C(n, k) = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.
Calculating Probability of Exactly 7 Defective Bulbs (P(X=7))
For $k=7$:
$C(8, 7) = \frac{8!}{7!(8-7)!} = \frac{8!}{7!1!} = 8$
$P(X=7) = C(8, 7) \times (0.1)^7 \times (0.9)^{(8-7)}$
$P(X=7) = 8 \times (0.1)^7 \times (0.9)^1$
$P(X=7) = 8 \times 10^{-7} \times 0.9 = 7.2 \times 10^{-7}$
To express this with a denominator of $10^8$, we write: $7.2 \times 10^{-7} = 72 \times 10^{-8}$.
Calculating Probability of Exactly 8 Defective Bulbs (P(X=8))
For $k=8$:
$C(8, 8) = \frac{8!}{8!(8-8)!} = \frac{8!}{8!0!} = 1$
$P(X=8) = C(8, 8) \times (0.1)^8 \times (0.9)^{(8-8)}$
$P(X=8) = 1 \times (0.1)^8 \times (0.9)^0$
$P(X=8) = 1 \times 10^{-8} \times 1 = 10^{-8}$
This can be written as $\frac{1}{10^8}$.
Total Probability P(X >= 7)
Summing the probabilities for exactly 7 and exactly 8 defective bulbs:
$P(X \ge 7) = P(X=7) + P(X=8)$
$P(X \ge 7) = (72 \times 10^{-8}) + (1 \times 10^{-8})$
$P(X \ge 7) = (72 + 1) \times 10^{-8}$
$P(X \ge 7) = 73 \times 10^{-8}$
$P(X \ge 7) = \frac{73}{10^8}$
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at random is tossed and head turns up. If the probability that the drawn coin was unbiased,
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