Sue goes through 4 sets of traffic lights each morning to work. there is a 5/8 chance that she will be caught by each of the lights. if she is stopped by 3 or more of the lighs then she is late for work. what is the probability that she will be late?

If there is an AND...AND situation we MULTIPLY the chances

Since the green light may be anywhere in the sequence (with the same probability) we may multiply the chance of just one green light by 4:

So here we have 5 possibilities of coming late. This is an EITHER...OR situation so we ADD probabilities:

The probability that Sue will be late for work is the sum of the probabilities that she will be stopped by 3, 4, or all 5 traffic lights.

Let's calculate each of these probabilities:

1. Probability of being stopped by exactly 3 lights: [\left(\frac{5}{8}\right)^3 \times \left(\frac{3}{8}\right)^1 \times \binom{4}{3} = \frac{405}{2048}]

2. Probability of being stopped by exactly 4 lights: [\left(\frac{5}{8}\right)^4 \times \left(\frac{3}{8}\right)^0 \times \binom{4}{4} = \frac{625}{4096}]

3. Probability of being stopped by all 5 lights: [\left(\frac{5}{8}\right)^5 \times \left(\frac{3}{8}\right)^0 \times \binom{4}{5} = \frac{3125}{32768}]

Therefore, the total probability of being late for work is: [\frac{405}{2048} + \frac{625}{4096} + \frac{3125}{32768} = \frac{369}{512} \approx 0.7207]

So, the probability that Sue will be late for work is approximately 0.7207 or 72.07%.