A fortune teller has numbers from $1 - 7$ in his box. He draws $3$ numbers from it. What is the probability that the number picked is of alternate order as odd-even-odd or even-odd-even?

1/14

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2/7

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9/14

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None of the above.

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Solution

The correct option is B
2/7

Sample Space: $7 \times 6 \times 5 = 210$

odd-even-odd= $4 \times 3 \times 3 = 36$

even-odd-even= $3 \times 4 \times 2 = 24$

Total ways: $60$

$P (E) = \frac{60}{210} = \frac{2}{7}$

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A fortune teller has numbers from 1−7 in his box. He draws 3 numbers from it. What is the probability that the number picked is of alternate order as odd-even-odd or even-odd-even?

The correct option is B 2/7 Sample Space: 7×6×5=210 odd-even-odd= 4×3×3=36 even-odd-even=3×4×2=24 Total ways: 60 P(E)=60210=27

A fortune teller has numbers from $1 - 7$ in his box. He draws $3$ numbers from it. What is the probability that the number picked is of alternate order as odd-even-odd or even-odd-even?

The correct option is B
2/7

Sample Space: $7 \times 6 \times 5 = 210$

odd-even-odd= $4 \times 3 \times 3 = 36$

even-odd-even= $3 \times 4 \times 2 = 24$

Total ways: $60$

$P (E) = \frac{60}{210} = \frac{2}{7}$