Question

10 cards are numbered from 1 to 10 and placed in a box. One card is selected at random and is not replaced. Another card is then randomly selected. What is the probability of selecting two numbers that are less than 6?
A. 2/9
B. 5/18
C. 1/5
D. 1/4

Answer 1

Answer:

Option A:    $\frac{\textbf{2}}{\textbf{9}}$

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 = $\frac{number \hspace{1mm} of \hspace{1mm} cards \hspace{1mm} less \hspace{1mm} than \hspace{1mm} 6}{total \hspace{1mm} number \hspace{1mm} of \hspace{1mm} cards}$

$= \frac{5}{10}$

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

$= \frac{4}{9}$

Since, both draw 1 and draw 2 are happening we multiply the two probabilities. We get

$\textbf{P} \hspace{1mm} \textbf{=} \hspace{1mm} \frac{\textbf{5}}{\textbf{10}} \hspace{1mm} \times \hspace{1mm} \frac{\textbf{4}}{\textbf{9}}$

$\therefore P = \frac{\textbf{2}}{\textbf{9}}$

Hence, OPTION A is the required answer.