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OpenStudy (anonymous):

Three quarters of commuters drive alone. What's the probability p that among 200 commuter vehicles chosen at random 150 or more are occupied by just one person?

OpenStudy (anonymous):

I'm not forsure on these questions...

OpenStudy (p0sitr0n):

3/4 = alone 1/4 = not alone 50/200 are not alone 50/200 = 1/4 same proportion, then maybe 100%?

OpenStudy (anonymous):

out of 200 50 drive alone 150 drive with someone 50/200=.25 so 25% probability

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