Question

You draw one card from a 52-card deck. Then the card is replaced in the deck and the deck is shuffled, and you draw again. Find the probability of drawing a diamond each time.

Answer 1

Solution:

Given:

A 52-card deck

There are four suits in a standard deck of cards, Clubs, Hearts, Spades, and Diamonds.

There are 13 diamond cards.

Hence,

$\begin{gathered} \text{Diamond cards = 13} \\ \text{Total cards = 52} \end{gathered}$

Probability is calculated by;

$\text{Probability}=\frac{n\text{ umber of required outcomes}}{n\text{ umber of total or possible outcomes}}$

Thus, the probability of drawing a diamond on the first draw is;

$\begin{gathered} \text{Probability of drawing a diamond}=\frac{n\text{ umber of diamond cards}}{\text{total number of cards}} \\ \text{Probability of drawing a diamond}=\frac{13}{52} \\ \text{Probability of drawing a diamond}=\frac{1}{4} \\ P(D_1)=\frac{1}{4} \end{gathered}$

Since two draws are made with replacement, the cards are completed back again before the next draw.

Hence, the probability of drawing a diamond on the second draw is;

$\begin{gathered} \text{Probability of drawing a diamond}=\frac{n\text{ umber of diamond cards}}{\text{total number of cards}} \\ \text{Probability of drawing a diamond}=\frac{13}{52} \\ \text{Probability of drawing a diamond}=\frac{1}{4} \\ P(D_2)=\frac{1}{4} \end{gathered}$

Therefore, the probability of drawing a diamond each time;

$\begin{gathered} P(D_1D_2)=\frac{1}{4}\times\frac{1}{4} \\ P(D_1D_2)=\frac{1}{16} \end{gathered}$