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?
