51- Dez cartas são extraídas aleatoriamente de um baralho usual de 52 cartas. Qual a probabilidade de todas serem de copas ?

10/52 tendo em vista que o baralho tenha 10 cartas de copa.

\[P=\dfrac{13}{52}\cdot\dfrac{12}{51}\cdots\dfrac{4}{43}\]

Outra forma de escrever, usando operação fatorial, é:

\[P=\left.{\dfrac{13!}{3!}}\right/{\dfrac{52!}{42!}}\]

Vamos simplificar aos poucos expandindo o fatorial e simplificando os fatores:

\[\eqalign{P&=\left.{\dfrac{\require{cancel}\cancel{13}\cdot12!}{3\cdot2!}}\right/{\dfrac{\require{cancel}\cancelto{4}{52}\cdot51!}{42\cdot41!}}\cr &=\left.{\dfrac{\require{cancel}\cancel{12}\cdot11!}{\require{cancel}\cancel3\cdot2!}}\right/{\dfrac{\require{cancel}\cancel4\cdot51!}{42\cdot41!}}\cr &=\left.{\dfrac{11\cdot10!}{\require{cancel}\cancel2\cdot1!}}\right/{\dfrac{51\cdot50!}{\require{cancel}\cancelto{21}{42}\cdot41!}}\cr &=11\cdot10!\left/{\dfrac{\require{cancel}\cancelto{17}{51}\cdot50!}{\require{cancel}\cancelto{7}{21}\cdot41!}}\right.\cr &=11\cdot\require{cancel}\cancelto{2}{10}\cdot9!\left/{\dfrac{17\cdot\require{cancel}\cancel{50}\cdot49!}{7\cdot41\cdot\require{cancel}\cancelto{4}{40}\cdot39!}}\right.\cr &=11\cdot\require{cancel}\cancel2\cdot9!\left/{\dfrac{17\cdot\require{cancel}\cancelto{7}{49}\cdot\require{cancel}\cancelto{6}{48}\cdot47!}{\require{cancel}\cancel7\cdot41\cdot\require{cancel}\cancel4\cdot39!}}\right.\cr &=11\cdot\require{cancel}\cancelto39\cdot8!\left/{\dfrac{17\cdot7\cdot\require{cancel}\cancel6\cdot47!}{41\cdot39\cdot\require{cancel}\cancelto{19}{38}\cdot37!}}\right.\cr &=11\cdot3\cdot8\cdot\require{cancel}\cancel7\cdot6!\left/{\dfrac{17\cdot\require{cancel}\cancel7\cdot47\cdot46!}{41\cdot39\cdot19\cdot37!}}\right.\cr &=11\cdot3\cdot\require{cancel}\cancelto48\cdot6!\left/{\dfrac{17\cdot47\cdot\require{cancel}\cancelto{23}{46}\cdot45!}{41\cdot39\cdot19\cdot37!}}\right.\cr &=11\cdot\require{cancel}\cancel3\cdot4\cdot\require{cancel}\cancelto26\cdot5!\left/{\dfrac{17\cdot47\cdot23\cdot\require{cancel}\cancelto5{45}\cdot44\cdot43!}{41\cdot39\cdot19\cdot37\cdot36\cdot35!}}\right.\cr &=11\cdot\require{cancel}\cancel4\cdot2\cdot\require{cancel}\cancel5\cdot4!\left/{\dfrac{17\cdot47\cdot23\cdot\require{cancel}\cancel5\cdot\require{cancel}\cancelto{11}{44}\cdot43!}{41\cdot39\cdot19\cdot37\cdot36\cdot35!}}\right.\cr &=\require{cancel}\cancel{11}\cdot\require{cancel}\cancel2\cdot4\cdot\require{cancel}\cancel3\cdot2\cdot1!\left/{\dfrac{17\cdot47\cdot23\cdot\require{cancel}\cancel{11}\cdot43\cdot\require{cancel}\cancelto{7}{42}\cdot41!}{41\cdot39\cdot19\cdot37\cdot36\cdot35!}}\right.\cr &=\require{cancel}\cancel{4\cdot2}\left/{\dfrac{17\cdot47\cdot23\cdot43\cdot7\cdot41\cdot\require{cancel}\cancelto5{40}\cdot\require{cancel}\cancel{39}\cdot38!}{41\cdot\require{cancel}\cancel{39}\cdot19\cdot37\cdot36\cdot35!}}\right.\cr &=1\left/{\dfrac{17\cdot47\cdot23\cdot43\cdot7\cdot41\cdot5\cdot38\cdot\require{cancel}\cancel{37\cdot36\cdot35!}}{41\cdot19\cdot\require{cancel}\cancel{37\cdot36\cdot35!}}}\right.\cr &=\dfrac{\require{cancel}\cancel{41}\cdot\require{cancel}\cancel{19}}{17\cdot47\cdot23\cdot43\cdot7\cdot\require{cancel}\cancel{41}\cdot5\cdot\require{cancel}\cancelto2{38}}\cr &=\dfrac{1}{17\cdot47\cdot23\cdot43\cdot7\cdot10}}\]

Então:

\[\boxed{P=\dfrac1{55314770}\approx1,8\cdot10^{-8}}\]