Question
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 <I>all</I> of these events will occur, you need to <I>multiply</I> 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
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 <I>all</I> of these events will occur, you need to <I>multiply</I> 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