Question

A box contains $9$ tickets numbered $1$ to $9$ inclusive. If $3$ tickets are drawn from the box without replacement, 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

The correct option is D

$\frac{5}{18}$

Explanation for the correct option:

Step 1: Find the number of possible outcomes.

We have been given that, a box contains $9$ tickets numbered $1$ to $9$ inclusive.

We need to find the probability, if $3$ tickets are drawn from the box without replacement, then they are alternatively either {odd, even odd} or {even, odd, even} .

Since a box contains $9$ tickets,

$\Rightarrow n\left(S\right)=9$

Step 2: Find the probability that three tickets were drawn in the order {odd, even, odd}

Even Numbered tickets $=\{2,4,6,8\}$

Odd number tickets $=\{1,3,5,7,9\}$

The probability that the first ticket drawn is odd = $\frac{5}{9}$

The total number of tickets left $=8$

So, the probability that the second ticket drawn is even $=\frac{4}{8}$

The total number of tickets left $=7$

The number of odd tickets left $=4$

So, the probability that the third ticket drawn is odd $=\frac{4}{7}$

Hence, the probability that three tickets are drawn in the order {odd, even, odd} would be,

$$\begin{gathered} =\frac{5}{9}\times \frac{4}{8}\times \frac{4}{7} \\ =\frac{10}{63} \end{gathered}$$

Step 3: Find the probability that three tickets were drawn in the order {even, odd, even}

Even Numbered tickets $=\{2,4,6,8\}$

Odd number tickets $=\{1,3,5,7,9\}$

The probability that the first ticket drawn is even $=\frac{4}{9}$

The number of tickets left $=8$

So, the probability that the second ticket drawn is odd $=\frac{5}{8}$

The number of tickets left $=7$

The number of even tickets left$=3$

So, the probability that the third ticket drawn is even$=\frac{3}{7}$

Hence, the probability that three tickets are drawn in the order {odd, even, odd} would be,

$$\begin{gathered} =\frac{4}{9}\times \frac{5}{8}\times \frac{3}{7} \\ =\frac{5}{42} \end{gathered}$$

Step 4: Find the required probability.

The probability that the three tickets are drawn either (odd, even, odd) or (even, odd, even) would be,

$$\begin{gathered} =\frac{10}{63}+\frac{5}{42} \\ =\frac{\mathbf{5}}{\mathbf{18}} \end{gathered}$$

Therefore, option (D) is the correct answer.