Question

A box contains $7$ tickets, numbered from $1$ to $7$ inclusive. If $3$ tickets are drawn from the box without replacement, one at a time, determine the probability that they alternatively either odd-even-odd or even -odd - even.

• A. $\frac{3}{7}$
• B. $\frac{5}{7}$
• C. $\frac{1}{7}$
• D. $\frac{2}{7}$

Answer 1

The correct option is D $\frac{2}{7}$

For odd-even-odd=

$P\left(A\right)=\frac{4}{7}\times \frac{3}{6}\times \frac{3}{5}=\frac{6}{35}$

For even-odd-even=

$P\left(B\right)=\frac{3}{7}\times \frac{4}{6}\times \frac{2}{5}=\frac{4}{35}$

$\therefore P=P\left(A\right)+P\left(B\right)=\frac{6}{35}+\frac{4}{35}=\frac{10}{35}=\frac{2}{7}$