Question

Nine cards are labelled \[0,1,2,3,4,5,6,7,8\]. Two cards are drawn at random and put on a table in a successive order, and then the resulting number is read, say, 07 (seven), 14 (fourteen) and so on. The probability that the number is even, is?
A. 5/9
B.4/9
C. 1/2
D. 2/3

Answer 1

Hint:The favourable outcomes for the event that the drawn card number is even is the sum of the favourable outcomes that both the cards are even and the favourable outcomes that the drawn first card is odd and the second card is even. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given nine cards are \[0,1,2,3,4,5,6,7,8\]
Out of these even cards are \[0,2,4,6,8\]
And odd cards are \[1,3,5,7\]
So, number of even cards = 5
And number of odd cards = 4
We know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
The total number of outcomes of drawing two cards is \[{}^9{C_2}\].
The favourable outcomes for the event that the drawn card number is even is the sum of the favourable outcomes that both the cards are even and the favourable outcomes that the drawn first card is odd and the second card is even.
Let \[A\] be the event that both the cards drawn are even.
So, number of favourable outcomes for first card = 5
And number of favourable outcomes of for second card = 4
Thus, the total number of favourable outcomes of even \[A\] \[ = 5 \times 4 = 20\]
Let \[B\] be the event that the drawn first card is odd and the second card is even.
So, number of favourable outcomes for first card = 4
And number of favourable outcomes for second card = 5
Thus, the total number of favourable outcomes for event \[B\] \[ = 4 \times 5 = 20\]
Hence the favourable outcomes for the event that the drawn card number is even = number of favourable outcomes of even \[A\] + number of favourable outcomes for event \[B\]
Thus, the required favourable outcomes \[ = 20 + 20 = 40\]
Therefore, the required probability \[ = \dfrac{{40}}{{{}^9{C_2}}} = \dfrac{{40}}{{72}} = \dfrac{5}{9}\]
Thus, the correct option is A. \[\dfrac{5}{9}\]
Note: The probability of an event \[E\] is always greater than or equal to zero and less than or equal to one i.e., \[0 \leqslant P\left( E \right) \leqslant 1\]. The number of outcomes is always greater than the number of favourable outcomes.