Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, 04-BS-16 Discrete Mathematics, May 2014. Closed book, no aids. The exam instructs "answer 10 of 12 questions"; every question is answered below as a complete study resource.
Reference texts: Rosen, Discrete Mathematics and Its Applications, 7th ed. (logic, induction, combinatorics, probability, relations, graph theory).
Given. A standard 52-card deck (13 ranks, 4 suits per rank), dealt 5-card hands with no replacement, as tabulated above.
Find. $P(\text{four of a kind})$, $P(\text{flush})$, $P(\text{two pair})$, $P(\text{three of a kind})$, $P(\text{full house})$.
Approach. Every probability is (favourable hands)/$\binom{52}{5}$; count favourable hands with the multiplication rule, choosing ranks first (unordered, $\binom{13}{\cdot}$) then suits within each chosen rank ($\binom{4}{\cdot}$).
a) Four of a kind. Choose the rank for the quad ($\binom{13}{1}$), take all 4 suits ($\binom{4}{4}=1$), then choose the 5th card from the remaining 48 cards:
$$\binom{13}{1}\binom{4}{4}\binom{48}{1}=13\cdot1\cdot48=624 \qquad P=\frac{624}{2{,}598{,}960}\approx 0.000240$$
$\boxed{P(\text{four of a kind})=\dfrac{624}{2{,}598{,}960}\approx 2.40\times10^{-4}}$
b) Flush (incl. straight flush). "All five of the same suit" is choose a suit ($\binom{4}{1}$) then any 5 of its 13 ranks ($\binom{13}{5}$) — this count already includes straight flushes since the question says "may or may not be in sequence":
$$\binom{4}{1}\binom{13}{5}=4\cdot1287=5{,}148 \qquad P=\frac{5{,}148}{2{,}598{,}960}\approx 0.00198$$
$\boxed{P(\text{flush})=\dfrac{5{,}148}{2{,}598{,}960}\approx 1.98\times10^{-3}}$
c) Two pair. Choose the 2 paired ranks ($\binom{13}{2}$), 2 suits for each ($\binom{4}{2}$ twice), then the 5th "kicker" card from the 44 remaining ranks' cards:
$$\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{44}{1}=78\cdot6\cdot6\cdot44=123{,}552 \qquad P=\frac{123{,}552}{2{,}598{,}960}\approx 0.0475$$
$\boxed{P(\text{two pair})=\dfrac{123{,}552}{2{,}598{,}960}\approx 4.75\times10^{-2}}$
d) Three of a kind (not a full house). Choose the triple's rank ($\binom{13}{1}$) and 3 of its suits ($\binom{4}{3}$), then choose 2 *different* ranks for the other two cards ($\binom{12}{2}$) each in one of 4 suits:
$$\binom{13}{1}\binom{4}{3}\binom{12}{2}\cdot4\cdot4=13\cdot4\cdot66\cdot16=54{,}912 \qquad P=\frac{54{,}912}{2{,}598{,}960}\approx 0.0211$$
$\boxed{P(\text{three of a kind})=\dfrac{54{,}912}{2{,}598{,}960}\approx 2.11\times10^{-2}}$
e) Full house. Choose the triple's rank ($\binom{13}{1}$) and 3 suits ($\binom{4}{3}$), then the pair's rank from the remaining 12 ranks ($\binom{12}{1}$) and 2 of its suits ($\binom{4}{2}$):
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}=13\cdot4\cdot12\cdot6=3{,}744 \qquad P=\frac{3{,}744}{2{,}598{,}960}\approx 0.00144$$
$\boxed{P(\text{full house})=\dfrac{3{,}744}{2{,}598{,}960}\approx 1.44\times10^{-3}}$
