P(\text{at least 2 flows}) = 0.08 + 0.12 + 0.18 + 0.12 = 0.5
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- $ P(\overline{A} \cap B \cap C) = (1 - 0.4) \cdot 0.5 \cdot 0.6 = 0.6 \cdot 0.5 \cdot 0.6 = 0.18 $
- $ P(A \cap B \cap C) = 0.4 \cdot 0.5 \cdot 0.6 = 0.12 $
- Now summing these:
- Thus, the probability that at least two channels allow sufficient flow is:
- \boxed{0.5}
- Question: A science communicator is filming a video and plans to use 3 different animations, each independently having a 75% chance of being accepted by the editing team. What is the probability that fewer than two animations are accepted?
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Let $ x = \frac{a}{b} $, then $ x + \frac{1}{x} = 5 \Rightarrow x^2 + \frac{1}{x^2} = 23 $, so:
x^4 + 1 = x^4 + \frac{1}{x^4} \cdot x^4 = x^4 + 1 \Rightarrow \text{Not helpful}
Instead, use:
\frac{x^2 + 1}{x^2 - 1} + \frac{x^2 - 1}{x^2 + 1} = \frac{(x^2 + 1)^2 + (x^2 - 1)^2}{(x^2 - 1)(x^2 + 1)} = \frac{2x^4 + 2}{x^4 - 1} = 2 \cdot \frac{x^4 + 1}{x^4 - 1}
Let’s compute $ x^4 + 1 $, $ x^4 - 1 $:
We know $ x^2 + \frac{1}{x^2} = 23 \Rightarrow x^4 + \frac{1}{x^4} = 527 $, so:
x^4 + 1 = x^4 + \frac{1}{x^4} \cdot x^4 = x^4 + 1 \Rightarrow \text{Still not helpful}
Let’s try plugging values:
Let $ x = \frac{5 + \sqrt{21}}{2} $, then $ x + \frac{1}{x} = 5 $. Compute numerically:
x = \frac{5 + \sqrt{21}}{2} \approx \frac{5 + 4.583}{2} = \frac{9.583}{2} \approx 4.7915