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I have a set of 100 numbers (from 1 to 100), from which a selector randomly picks numbers. What is the probability, that the selector will pick numbers under 95 (equal or less) three times consecutive?

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    $\begingroup$ What is the chance the first number is under $95$? Do you count $95?$ Then use the multiplication principle to find the chance for all three. You have told us the draws are independent. $\endgroup$ – Ross Millikan Dec 31 '18 at 17:05
  • $\begingroup$ I do count 95, so it's equal or less than 95. $\endgroup$ – Ruham Dec 31 '18 at 17:06
  • $\begingroup$ Once a number is picked, does it remain available to be picked again? Or is it like removing cards from a deck? Also, what do you mean by "consequentially"? Do you mean consecutive (three in a row)? $\endgroup$ – timtfj Dec 31 '18 at 23:47
  • $\begingroup$ It does remain in the batch, yes, consecutive draws of numbers from 1 to 100 equal or less than 95. $\endgroup$ – Ruham Jan 1 '19 at 8:45
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I assume the obvious that the selector picks numbers with equal probability (or chances); also that it leaves those numbers there for the next pick. You essentially want it to pick one of the first 95 out of the 100 numbers. This gives you the selector's chances for one pick. Then you want it to do it 3 times in a row.

Maybe you know a similar example you can use: if you throw a coin and you know the probability of getting heads once, what is the probability of getting heads twice? This is easy to compute because you can count all the possible events for 2 throws. Think about 3 throws. You'll see how the probability for 1 pick gives the probability of 3 independent picks.

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  • $\begingroup$ Thanks, but I don't think this works with heads/tails. In that case, the probability of three draws is 1/8, but in my case, I have to take into account the chance of drawing a number less than the given. So if once it's 95%, I can't calculate what is the overall probability of three consequential 95% chance draws. $\endgroup$ – Ruham Dec 31 '18 at 17:38
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    $\begingroup$ Flip a coin. Probability of 1 draw is 1/2, for 2 draws is 1/4, for 3 draws 1/8, 4 draws at 1/16 etc. These derive from the initial 1 draw probability of 1/2. Now you have to hit one of those 95 numbers. Probability for one pick is 95/100. What about 2 or 3 picks? :) It should be the same situation in terms of probability. $\endgroup$ – Ferred Dec 31 '18 at 21:09
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    $\begingroup$ I guess than it would be 95/100 * 95/100 * 95/100 $\endgroup$ – Ruham Jan 1 '19 at 8:49
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    $\begingroup$ @Ruham Yup, that's right $\endgroup$ – Sauhard Sharma Jan 1 '19 at 9:38

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