10 of the numbers are even. 4 more are odd multiples of 3. The probability of landing on even or a multiple of 3 is (10+4)/21 = 2/3.
Answer:
7/8
Step-by-step explanation:
Answer:
about 0.254
Step-by-step explanation:
The light is red for 0.3 of the period, so that is the probability one car is stopped. Probability 3 cars are stopped and 5 are not is 0.3^3·0.7^5, about 0.004538. In the group of 8 cars, there are 56 different ways 3 of the cars can be stopped, so your overall probability could be 56·0.004538 ≈ 0.254.
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<em>Comment on the question</em>
Many factors go into a driver's decision to stop at a light. Many factors go into the distribution of arrival times at a light. Here, the problem is only tractable if we assume that cars arrive at the light individually and at random times with respect to the light's fixed 50-second cycle. (This is possibly the case only early in the morning hours when traffic is at its lightest (not associated with bar closings or night shift changes).)
