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} of {even, odd, even} is

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

Answer 1

The correct option is C $\frac{5}{18}$

First find probability that the $3$ tickets drawn are ( odd, even, odd )

There are $9$ tickets numbered $1$ to $9$

So, initially there are $5$ odd numbered tickets and $4$ even numbered.

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

Now, the second ticket is drawn without replacement. There are $8$ tickets in the box, out of which $4$ are odd numbered and $4$ are even.

The probability that second ticket drawn is even numbered $=\frac{4}{8}$

Now, the third ticket is drawn without replacement

Probability that third ticket drawn is odd numbered $=\frac{4}{7}$

probability that the $3$ tickets drawn are ( odd, even, odd )

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

Second case,

the $3$ tickets drawn are ( even, odd, even )

There are $9$ tickets numbered $1$ to $9$

So, initially, there are $5$ odd numbered tickets and $4$ even numbered.

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

Now, the second ticket is drawn without replacement. There are $8$ tickets in the box, out of which $5$ are odd numbered and $3$ are even.

The probability that the second ticket was drawn is odd numbered $=\frac{5}{8}$

Now, the third ticket is drawn without replacement

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

The probability that the $3$ tickets drawn are ( even, odd, even )

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

Probability that the $3$ tickets drawn are either ( even, odd, even ) or

( odd, even, odd ) $=\frac{10}{63}+\frac{5}{42}=\frac{5}{18}$