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.
What is the probability in fractions that at least one green is chosen?
What is the probability in fractions that at least one green is chosen?
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ilovekasha6 -
The "best" way to solve an "at least" probability is always take the complement.
P(at least one green is chosen) = 1 - P(no green), so let's find the P(no green):
P(no green) = 4C2 / 9C2 = 6 / 36 = 1/6
Finally, find 1 - P(no green)
1 - P(no green) = 1 - 1/6 = 5/6
Hope that helped
The "best" way to solve an "at least" probability is always take the complement.
P(at least one green is chosen) = 1 - P(no green), so let's find the P(no green):
P(no green) = 4C2 / 9C2 = 6 / 36 = 1/6
Finally, find 1 - P(no green)
1 - P(no green) = 1 - 1/6 = 5/6
Hope that helped