SOLUTION: Suppose that you have 9 cards. 5 are green and 4 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 4 yellow cards are numbered 1, 2, 3, and 4. The cards are well sh
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-> SOLUTION: Suppose that you have 9 cards. 5 are green and 4 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 4 yellow cards are numbered 1, 2, 3, and 4. The cards are well sh
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Question 1094068: Suppose that you have 9 cards. 5 are green and 4 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 4 yellow cards are numbered 1, 2, 3, and 4. The cards are well shuffled. Suppose that you randomly draw two cards, one at a time, and without replacement.
• G1 = first card is green
• G2 = second card is green
Enter the probability as a fraction.
P(at least one green) = Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! The easiest way to solve is to first compute P(no green).
Then the probability of at least one green is
P(at least one green) = 1 - P(no green)
P(no green) = 4/9*3/8 = 12/72 = 1/6
Thus P(at least one green) = 1 - 1/6 = 5/6
