A bag contains 9 red marbles and 3 green marbles. How can you find the probability of drawing first one green marble and then one red marble if you do not replace the first marble?

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A bag contains 9 red marbles and 3 green marbles. How can you find the probability of drawing first one green marble and then one red marble if you do not replace the first marble?

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Answer 1 · 2015-11-02T09:11:22

P(First Marble Green and Second Marble Red) $= \frac{9}{44}$ $=9/44$

Explanation:

Probability for the first marble to be green $= \frac{3}{9 + 3} = \frac{3}{12} = \frac{1}{4}$ $=3/(9 +3)=3/12=1/4$

Probability for the second marble to be red (marble taken first is not replaced) $= \frac{9}{9 + (3 - 1)} = \frac{9}{9 + 2} = \frac{9}{11}$ $=9/(9+(3-1))=9/(9+2)=9/11$

P(First Marble Green and Second Marble Red) $= \frac{1}{4} \times \frac{9}{11} = \frac{9}{44}$ $=1/4 xx 9/11=9/44$

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A bag contains 9 red marbles and 3 green marbles. How can you find the probability of drawing first one green marble and then one red marble if you do not replace the first marble?

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