Consider a bridge hand (that is, draw 13 cards at random and without replacement from a regular, 52-card deck). Find the following probabilities: are. Exactly 6 spades, 4 hearts, 2 diamonds, and 1 club

Answer:The probability of selecting exactly 6 spades, 4 hearts, 2 diamonds, and 1 club is [tex]P=\frac{^{13}C_{6}\times ^{13}C_{4}\times ^{13}C_{2}\times ^{13}C_{1}}{^{52}C_{13}}[/tex].Step-by-step explanation:Total number of cards in a regular deck of cards = 52Total number of cards of each suit (spades, hearts, diamonds, club) = 13Total ways of selecting r cards from total n cards is[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]Total ways of selecting 13 cards from total 52 cards is[tex]\text{Total outcomes}=^{52}C_{13}[/tex]Total ways of selecting exactly 6 spades, 4 hearts, 2 diamonds, and 1 club is[tex]\text{Favorable outcomes}=^{13}C_{6}\times ^{13}C_{4}\times ^{13}C_{2}\times ^{13}C_{1}[/tex]The probability of selecting exactly 6 spades, 4 hearts, 2 diamonds, and 1 club is[tex]P=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex][tex]P=\frac{^{13}C_{6}\times ^{13}C_{4}\times ^{13}C_{2}\times ^{13}C_{1}}{^{52}C_{13}}[/tex]Therefore the probability of selecting exactly 6 spades, 4 hearts, 2 diamonds, and 1 club is [tex]P=\frac{^{13}C_{6}\times ^{13}C_{4}\times ^{13}C_{2}\times ^{13}C_{1}}{^{52}C_{13}}[/tex].