Answer to Question #1737 in Statistics and Probability for Tim
Probability for 1st man men to get black bean
P1 (black) = 1000/1001.
Probability for 1st man men to get green bean
P1 (green) = 1/1001.
2. 999 black beans, 1 green bean(we suppose that beans we don’t put beans back).
Probability for 2nd man men to get black bean
P2 (black) = 999/1000
P2 (green) = P1 * 1/1000 = 1000/1001 * 1/1000
3. 998 black beans ,1 green
P3 (black) = 998/999
P3 (green) = P1 (g)P2 (g) * 1/1000 = 1000/1001 * 999/1000 * 1/999
n. For 999th man probability will be
P999 (green) = P1 (g) P2 (g)…P997 (g) P998 (g) * 1/1000 = 1000/1001 * 999/1000*…*3/4 * 2/3 * 1/2
If after a draw beans go back to jar probability for everyone will be the same and will be equal
P(black) = 1000/1001.
P(green) = 1/1001.
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Thank you for your question. In this experiment the output of a previous trial is very important, because if a person picks out a green bean, then probability, that other person will take out a green bean, equals zero. We use probability of intersection of events and conditional probability: P(C and D)=P(C)*P(D/C), where P(C)=P1, P(D/C)=1/1000.
if u don't mind i have a doubt
can i know why is P2 (green) = P1 * 1/1000 and not 1/1000 ??????