anonymous
  • anonymous
an urn contains 6 white and 9 blacks balls. if 4 balls are to be randomly selected without replacement, what is the probabilty that the first 2 selected are white and the last 2 black?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The probability that the first one is white is 2/5. The probability that the second is white, given that the first was, is 2/5*5/14. The chance the third ball is black, given the first two were white, is 2/5*5/14*9/13. The chance the first two are white, then the next two are black is 2/5*5/14*9/13*2/3
anonymous
  • anonymous
About 18/273, which is a little less than 7%.
anonymous
  • anonymous
here ,first find the probability to get first two selected ball is white. P(w)=(6C2/15C2). And then find the probability to get second two ball is black. P(b)=(9C2/13C2).. Here follow multiplication rule. Required probability=P(w)*P(b)...find it

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anonymous
  • anonymous
Required probability=18/273...
anonymous
  • anonymous
Nice approach. A little more sophisticated than my basic counting, but same answer.

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