College Algebra, Chapter 10, 10.4, Section 10.4, Problem 32

Three cards are randomly selected from a standard 52-card deck, one at a time, with each card replaced in the deck before the next one is picked. Find the probability of each event.

a.) All three cars are hearts.

b.) Exactly two of the cards are spades.

c.) None of the cards is a diamond.

d.) At least one of the cards is a club.

Recall that the formula for the binomial probability is given by

$C(n,r) p^r q^{n-r}$

In this case, we have $n =3$.

a.) For this event, the probability that the card drawn is heart is $\displaystyle \frac{13}{52} = \frac{1}{4}$. This gives the probability of success $p=0.25$ and probability of failure $q=1-p=0.75$. Thus, the probability that all three cards are heart (value $r=3$) is

$= C(3,3) (0.25)^3 (0.75)^{3-3}$

$= 0.0156$

b.) Similar from part (a), the probability that the card drawn is spade is $\displaystyle \frac{13}{52} = \frac{1}{4}$. Thus, the probability that exactly two of the cards are spades (value $r=2$) is

$= C(3,2) (0.25)^2 (0.75)^{3-2}$

$= 0.1406$

c.) For this event, the probability that the card drawn is diamond is $\displaystyle \frac{13}{52} = \frac{1}{4}$. This gives $p=0.25$ and $q=1-p=0.75$. Thus, the probability that all the cards is diamond (value $r=0$) is

$= C(3,0) (0.25)^0 (0.75)^{3-0}$

$= 0.4219$

d.) We know that the probability that the card drawn is club is $\displaystyle \frac{13}{52} = \frac{1}{4}$. To solve the probability if this event in more efficient manner, we get the complement of the probability that all three drawn cards are not clubs. In this case, we have $p=0.25, q=1-p=0.75, n=3$ and $r=0$. So we get

$
\begin{equation}
\begin{aligned}

=& 1- [C(3,0) (0.25)^0 (0.75)^{3-0}]
\\
\\
=& 1-0.4219
\\
\\
=& 0.5781

\end{aligned}
\end{equation}
$