A fortune teller has numbers from 1-7 in his box. He draws 3 numbers from it.

Discussion Forum : Probability I

Question -

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?

Options:

A .

\frac{1}{14}

B .

\frac{2}{7}

C .

\frac{9}{14}

D . None of the above.

Answer: Option B
:
B

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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