To determine the probability P(7 or more) in a binomial distribution, where $n = 9$ (number of trials) and $p = 0.18$ (probability of success), we use the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

The cumulative probability $P(X \geq 7)$ can be calculated as:

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9)$

Step 1: Formula Components

• $n = 9$, $p = 0.18$, and $q = 1 - p = 0.82$.
• The binomial coefficient is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Step 2: Calculate Individual Probabilities

For $X = 7$:

$P(X = 7) = \binom{9}{7} (0.18)^7 (0.82)^2$ $\binom{9}{7} = \frac{9!}{7! \cdot 2!} = 36, \quad (0.18)^7 \approx 2.797e{-6}, \quad (0.82)^2 = 0.6724$ $P(X = 7) = 36 \cdot 2.797e{-6} \cdot 0.6724 \approx 6.77e{-5}$

For $X = 8$:

$P(X = 8) = \binom{9}{8} (0.18)^8 (0.82)^1$ $\binom{9}{8} = \frac{9!}{8! \cdot 1!} = 9, \quad (0.18)^8 \approx 5.035e{-7}, \quad (0.82)^1 = 0.82$ $P(X = 8) = 9 \cdot 5.035e{-7} \cdot 0.82 \approx 3.72e{-6}$

For $X = 9$:

$P(X = 9) = \binom{9}{9} (0.18)^9 (0.82)^0$ $\binom{9}{9} = 1, \quad (0.18)^9 \approx 9.063e{-8}, \quad (0.82)^0 = 1$ $P(X = 9) = 1 \cdot 9.063e{-8} \cdot 1 \approx 9.063e{-8}$

Step 3: Add Probabilities

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9)$ $P(X \geq 7) \approx 6.77e{-5} + 3.72e{-6} + 9.063e{-8}$ $P(X \geq 7) \approx 7.15e{-5}$

Final Answer:

$P(X \geq 7) \approx 0.0001$

Let me know if you'd like a breakdown of any step! 😊

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Explore More:

1. How does the binomial distribution differ from normal distribution?
2. How do changes in $p$ affect $P(X \geq 7)$?
3. What happens when $n$ increases to 20 for the same $p$?
4. How would $P(X \leq 6)$ be calculated here?
5. Can we approximate this using the Poisson distribution?

Tip: Use cumulative binomial tables or a calculator to simplify repetitive calculations!