Answer:
Option A: [tex]$ \frac{\textbf{2}}{\textbf{9}} $[/tex]
Step-by-step explanation:
Given there are 10 cards viz: 1, 2, 3, 4, . . . , 10
We find the probability of drawing two cards less than six, without replacing the first card.
Draw 1:
There are 5 cards with value less than 6. 1, 2, 3, 4, 5
The total number of cards is 10.
The probability of the number being less than 6 = [tex]$ \frac{number \hspace{1mm} of \hspace{1mm} cards \hspace{1mm} less \hspace{1mm} than \hspace{1mm} 6}{total \hspace{1mm} number \hspace{1mm} of \hspace{1mm} cards} $[/tex]
[tex]$ = \frac{5}{10} $[/tex]
Draw 2:
We are again drawing a card without replacing the card that was drawn earlier. This makes the total number of cards 9.
Also, the number of cards less than 6 will now be: 4.
Therefore, probability of drawing a number less than 6 without replacing
[tex]$ = \frac{4}{9} $[/tex]
Since, both draw 1 and draw 2 are happening we multiply the two probabilities. We get
[tex]$ \textbf{P} \hspace{1mm} \textbf{=} \hspace{1mm} \frac{\textbf{5}}{\textbf{10}} \hspace{1mm} \times \hspace{1mm} \frac{\textbf{4}}{\textbf{9}} $[/tex]
[tex]$ \therefore P = \frac{\textbf{2}}{\textbf{9}} $[/tex]
Hence, OPTION A is the required answer.
