**–Â Two hearts are drawn in the first two drawings.Â **

**– The first draw was a heart, and the second draw was a club.**

The main objective of this **question** is to find the **probability** of** cards drawn** from the **deck**.

This question **uses** the concept of **probability**. Probability is a **branch** of **mathematics** that uses **numbers** to **describe** how likely it is that **something** will **happen** or that a **statement** is **true**.

## Expert Answer

a)Â We **know** that:

\[ \space P A \cap B \space = \space P ( A ) \space \times \space P ( B | A )Â \space = \space P ( B ) \space \times \space P ( A | b ) \]

**So**:

The** probability** of $ A $ is:

\[ \space P ( A ) \space = \space \frac{ 1 3 }{ 5 2 } \]

**And**:

\[ \space P( B | A ) space = \space \frac{ 1 2 }{ 51 } \]

**Substituting** the **values**, we get:

\[ \space = \space \frac{ 1Â 3 }{ 5 2 } \space \times \space \frac{Â 1 2 }{ 5 1 } \]

\[ \space = \space \frac{ 1 }{Â 1 7 } \]

b) We **know** that:

\[ \space P A \cap B \space = \space P ( A ) \space \times \space P ( B | A )Â \space = \space P ( B ) \space \times \space P ( A | b ) \]

**So**:

The **probability** of $ A $ is:

\[ \space P ( A ) \space = \space \frac{ 1 3 }{ 5 2 } \]

**And**:

\[ \space P( B | A ) space = \space \frac{ 1 3 }{ 51 } \]

**Substituting** the** values**, we get:

\[ \space = \space \frac{ 1Â 3 }{ 5 2 } \space \times \space \frac{Â 1 3 }{ 5 1 } \]

\[ \space = \space \frac{ 1 3Â }{Â 2 0 4 } \]

## Numerical Answer

The probability of t**wo hearts **being** drawn **in the** first two drawings is:**

\[ \space = \space \frac{ 1 }{Â 1 7 } \]

The probability that the**Â first draw **was a** heartÂ **and the** second draw **was a** clubÂ **is:

\[ \space = \space \frac{ 1 3Â }{Â 2 0 4 } \]

## Example

A regular** deck** of **cards** is used to** draw** two cards one after the otherÂ **without** replacing them. **Figure** out the chances of **drawing**. Find the** probability** that the two cards are **drawn** as **diamonds**.

We **know** that:

\[ \space P A \cap B \space = \space P ( A ) \space \times \space P ( B | A )Â \space = \space P ( B ) \space \times \space P ( A | b ) \]

**So**:

The** probability** of $ A $ is:

\[ \space P ( A ) \space = \space \frac{ 1 3 }{ 5 2 } \]

**And**:

\[ \space P( B | A ) space = \space \frac{ 1 2 }{ 51 } \]

**Substituting** the **values**, we get:

\[ \space = \space \frac{ 1Â 3 }{ 5 2 } \space \times \space \frac{Â 1 2 }{ 5 1 } \]

\[ \space = \space \frac{ 1 }{Â 1 7 } \]