 # Two cards are drawn successively and without replacement from an ordinary deck of playing cards Compute the probability of drawing –  Two hearts are drawn in the first two drawings.

– 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.

a)  We know that:

$\space P A \cap B \space = \space P ( A ) \space \times \space P ( B | A ) \space = \space P ( B ) \space \times \space P ( A | b )$

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:

$\space = \space \frac{ 1 3 }{ 5 2 } \space \times \space \frac{ 1 2 }{ 5 1 }$

$\space = \space \frac{ 1 }{ 1 7 }$

b) We know that:

$\space P A \cap B \space = \space P ( A ) \space \times \space P ( B | A ) \space = \space P ( B ) \space \times \space P ( A | b )$

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:

$\space = \space \frac{ 1 3 }{ 5 2 } \space \times \space \frac{ 1 3 }{ 5 1 }$

$\space = \space \frac{ 1 3 }{ 2 0 4 }$

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

$\space = \space \frac{ 1 }{ 1 7 }$

The probability that the first draw was a heart and the second draw was a club is:

$\space = \space \frac{ 1 3 }{ 2 0 4 }$

## Example

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

We know that:

$\space P A \cap B \space = \space P ( A ) \space \times \space P ( B | A ) \space = \space P ( B ) \space \times \space P ( A | b )$

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:

$\space = \space \frac{ 1 3 }{ 5 2 } \space \times \space \frac{ 1 2 }{ 5 1 }$

$\space = \space \frac{ 1 }{ 1 7 }$