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A bag contains 19 blue balls and 11 red balls. 8 balls are drawn from the bag one by one without replacement. What is the probability that the 8th ball drawn is red?

I am a bit confused so as to how to approach this problem. Bcoz I think, how can i calculate what is going to happen 8th draw, unless I have information about what has happen in the last 7 draws...? Plz explain it ...

Piyush Sawarkar
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  • Suppose you drew out all of the balls (i.e., a permutation). What proportion of the permutations are there with a red ball in the $8$th position? – Rushabh Mehta Jan 14 '20 at 16:09
  • The end result is that if you have a bag of items and you draw multiple times with or without replacement, the probability that the $k$'th draw is of a certain type will be exactly the same as the probability that the first draw is of that type. The answer here is $\frac{11}{30}$. The information of what the first however many draws were is entirely irrelevant to the problem. – JMoravitz Jan 14 '20 at 16:11
  • @JMoravitz I guess as per what you said it should be $\frac{11}{30}$ plz correct me if i am wrong.. – Piyush Sawarkar Jan 14 '20 at 16:15
  • @PeterSmith right., I had read 19 total balls, 8 of which red somehow... probably because $19=11+8$ – JMoravitz Jan 14 '20 at 16:16
  • Okay thank you... – Piyush Sawarkar Jan 14 '20 at 16:17

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