Question

A box contains $9$ tickets numbered $1$ to $9$ inclusive. If $3$ tickets are drawn from the box one at a time, the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is

• A. $\frac{5}{17}$
• B. $\frac{4}{17}$
• C. $\frac{5}{16}$
• D. $\frac{5}{18}$

Answer 1

Answer: (D)

No. of tickets = 9
No. of odd numbered tickets = 5
No. of even numbered tickets = 4
Required probability = P {odd, even, odd}
+ P (even, odd, even)
$=\frac{^{5}C_{1}}{^{9}C_{1}}\times\frac{^{4}C_{1}}{^{8}C_{1}}\times\frac{^{4}C_{1}}{^{7}C_{1}}\times\frac{^{4}C_{1}}{^{9}C_{1}}\times\frac{^{5}C_{1}}{^{8}C_{1}}\times\frac{^{3}C_{1}}{^{7}C_{1}} $
$=\frac{5}{9}\times\frac{4}{8}\times\frac{4}{7}\times \frac{4}{9}\times \frac{5}{8}\times \frac{3}{7}= \frac{5}{18} $