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wooden balls are numbered 1-75. Five balls are drawn at random. What is the probability of drawing 5 consecutive numbers if no replacement occurs.?

The first ball has a 1/75 chance of being selected. The second ball can be from either above or below the first, so it has a 2/74 chance of being chosen. The same process of selection continues through the fifth ball.

To find the probability that all of these events will occur, you need to multiply the probability of the individual events.

With this information, you should be able to answer your own question.

I hope this helps. Thanks for asking.

I don't think that's right PsyDAG. What if the first ball is #1 -- there is no number below. Second, what if you picked balls in the following order; 5,3,1,2,4 would this be counted as a "success" as the numbers are, when rearranged, consecutive.

I think this is a combinatorial problem. The number of ways do draw 5 balls from a bag of 75 is 75-choose-5 or (75!)/((75-5)!*(5!)) where ! denotes factorial. This translates to (75*74*73*72*71)/(1*2*3*4*5) = 17259390. Now then, there are exactly 70 5-consecutive-balls combinations that could be drawn. (1,2,3,4,5), (2,3,4,5,6), ... (71,72,73,74,75). So, the probability is 70/17259390 = something very small

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