Question

Ten cards, numbered $11$ to $20$, are placed in a bowl. Two cards are randomly drawn one by one with replacement. What is the probability of drawing a card with an even number then a card with a number greater than or equal to fifteen?

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{5}$
• D. $\frac{3}{10}$

Answer 1

The correct option is D $\frac{3}{10}$

Numbered cards in the bowl= $\text{{11, 12, 13, 14, 15, 16, 17, 18, 19, 20}}$
$\therefore$ Total number of cards in the bowl= $10$

Even numbered cards in the bowl= $\text{{12, 14, 16, 18, 20}}$
$\therefore$ Number of even cards in the bowl= $5$

Numbered cards in the bowl that are $\ge 15$= $\text{{15, 16, 17, 18, 19, 20}}$
$\therefore$ Number of cards in the bowl with number $\ge 15$= $6$

Suppose, drawing a card with an even number is event $\text{A}$ and drawing a card with a number greater than or equal to $15$ is event $\text{B}$.

Hence,
$\text{P(A)}=\frac{\text{Number of even numbered cards in bowl}}{\text{Total number of cards in bowl}}$

$\Rightarrow \text{P(A)}=\frac{5}{10}=\frac{1}{2}$

Next,
$\text{P(B)}=\frac{\text{Number of cards in bowl with number}\ge 15}{\text{Total number of cards in bowl}}$

$\Rightarrow \text{P(B)}=\frac{6}{10}=\frac{3}{5}$
(As first drawn card is replaced, total number of cards will not change)

Now, $\text{P(A and B) = P(A)}\times \text{P(B)}$
$\Rightarrow \text{P(A and B)}=\frac{1}{2}\times \frac{3}{5}=\frac{1\times 3}{2\times 5}=\frac{3}{10}$

$\therefore$ The probability of randomly drawing an even numbered card then a card with a number greater than or equal to fifteen is $\frac{3}{10}$.