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) $\dfrac{5}{{17}}$
B) $\dfrac{4}{{17}}$
C) $\dfrac{5}{{16}}$
D) $\dfrac{5}{{18}}$

Answer 1

The meaning of probability is basically the extent to which something is likely to happen. To find the probability of a single event to occur, first, we should know the total number of possible outcomes. Here, we are given 9 tickets (1 to 9). Out of these, 5 are odd tickets and 4 are even tickets. Then, we need to find the probability that the 3 tickets drawn are (odd-even-odd) and then 3 tickets drawn are (even-odd-even). Thus, we need to add both the probability to get the final output.

Complete step by step solution:
Given that, There are 9 tickets = {1, 2, 3, 4, 5, 6, 7, 8, 9}.
In that, Odd numbered tickets = {1, 3, 5, 7, 9} which means that there are a total 5 odd tickets and Even numbered tickets = {2, 4, 6, 8} which means that there are a total 4 even numbered tickets.
Here, we will find the probability in two different cases as below:

Case I: odd-even-odd
In this we need to find the first ticket drawn is odd, then even and last odd without any replacements.
So, the probability that the first ticket drawn is odd numbered $= \dfrac{5}{9}$ .
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.
So, The probability that the second ticket is drawn is even numbered $= \dfrac{4}{8}$ .
Now, the third ticket is drawn without replacement. There are 7 tickets in the box, out of which 4 are odd numbered and 3 are even.
So, The probability that the second ticket is drawn is odd numbered $= \dfrac{4}{7}$ .
Thus, the probability that the 3 tickets drawn (odd-even-odd) are
$= \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}$
$= \dfrac{{5 \times 2}}{{9 \times 7}}$
$= \dfrac{{10}}{{63}}$

Case II: even-odd-even
In this we need to find the first ticket drawn is even, then odd and last even without any replacements.
So, the probability that the first ticket drawn is even numbered $= \dfrac{4}{9}$ .
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.
So, The probability that the second ticket is drawn is odd numbered $= \dfrac{5}{8}$ .
Now, the third ticket is drawn without replacement. There are 7 tickets in the box, out of which 4 are odd numbered and 3 are even.
So, The probability that the second ticket is drawn is even numbered $= \dfrac{3}{7}$ .
Thus, the probability that the 3 tickets drawn (even-odd-even) are
$= \dfrac{4}{9} \times \dfrac{5}{8} \times \dfrac{3}{7}$
$= \dfrac{5}{{3 \times 2 \times 7}}$
$= \dfrac{5}{{42}}$

Now, the probability that the 3 tickets drawn are either (odd-even-odd) or (even-odd-even)
$= \dfrac{{10}}{{63}} + \dfrac{5}{{42}}$
$= \dfrac{{10(2) + 5(3)}}{{126}}$
$= \dfrac{{20 + 15}}{{126}}$
$= \dfrac{{35}}{{126}}$
$= \dfrac{5}{{18}}$

Hence, the probability that the 3 tickets drawn are either (odd-even-odd) or (even-odd-even) is $\dfrac{5}{{18}}$.

Note:
Probability means possibility. It deals with the occurrence of a random event. The value is expressed from zero to one. The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes. Sometimes students get mistaken for a favourable outcome with a desirable outcome.