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ohaa [14]
3 years ago
9

From 6 cards bearing the letters a, b, c, d, e, and f, two cards are drawn. Find the probability that at least one of the cards

drawn bears a vowel. Note: (a, b) and (b, a) are considered to be different elements in the sample space.
A) 2/15
B) 3/5
C) 5/9
Mathematics
2 answers:
alexgriva [62]3 years ago
7 0
We use the complement rule of probability.
P(at least one vowel) = 1 - P(no vowel)
To solve for P(no vowel) = (4/6)(3/5) = 2/5; since there are 4 consonants out of 6 letters and we take the letters without replacement.

P(at least one vowel) = 1 - 2/5 = 3/5.
The answer is B.
Tomtit [17]3 years ago
7 0
The answer is B just tried it and it is correct. Hope this helps.

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Answer:

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Answer:

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Step-by-step explanation:

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To meet the condition on the fifth day, there needs to be a yellow light. The probability that the condition is met on the first four days and on the fifth day will be (0.5 \times 0.5 \times 0.5 \times 0.5) \times 0.1 = 0.5^{4} \times 0.1.

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