Question #1737

1000 black beans in a a jar, plus 1 green bean. If 999 people draw beans from the jar, what is the probability of each one drawing the green bean? Need the answer and the equation that leads to the answer, please.

Expert's answer

1. 1000 black beans, 1 green bean.

Probability for 1st man men to get black bean

**P**_{1} (black) = 1000/1001.

Probability for 1st man men to get green bean

**P**_{1} (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

**P**_{2} (black) = 999/1000

P_{2} (green) = P_{1} * 1/1000 = 1000/1001 * 1/1000

3. 998 black beans ,1 green

**P**_{3} (black) = 998/999

P_{3} (green) = P_{1} (g)P_{2} (g) * 1/1000 = 1000/1001 * 999/1000 * 1/999

n. For 999th man probability will be

**P**_{999} (green) = P_{1} (g) P_{2} (g)…P_{997} (g) P_{998} (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.

Probability for 1st man men to get black bean

Probability for 1st man men to get green bean

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

P

3. 998 black beans ,1 green

P

n. For 999th man probability will be

If after a draw beans go back to jar probability for everyone will be the same and will be equal

P(green) = 1/1001.

## Comments

Assignment Expert18.11.14, 17:30Dear doubt.

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.

doubt17.11.14, 20:04if u don't mind i have a doubt

can i know why is P2 (green) = P1 * 1/1000 and not 1/1000 ??????

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