Respuesta :
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 = 52
Total number of cards of each suit (spades, hearts, diamonds, club) = 13
Total 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].
Answer:
The probability of getting exactly 6 spades,4 hearts,2 diamond and one club=[tex]\frac{\binom{13}{6}\times\binom{13}{4}\times \binom{13}{2}\times\binom{13}{1}}{\binom{52}{13}}[/tex]
Step-by-step explanation:
We are given that a bridge hand that draw 13 cards at random and without replacement from a regular 52 cards deck.
We have to find the probabilities
1.exactly 6 spades
We know that in one deck
Total number of spades=13
Total number of heart =13
Total number of diamond in a deck=13
Total number of clubs in a deck =13
Total number of cards=52
The probability of getting exactly 6 spades,4 hearts,2 diamond and one club=[tex]\frac{\binom{13}{6}\times\binom{13}{4}\times\binom{13}{2}\times\binom{13}{1}}{\binom{52}{13}}[/tex]
Probability,P(E)=[tex]\frac{number\;of\;favourable\;cases}{Total\;number\;of\;cases}[/tex]
The probability of getting exactly 6 spades,4 hearts,2 diamond and one club=[tex]\frac{\binom{13}{6}\times\binom{13}{4}\times \binom{13}{2}\times\binom{13}{1}}{\binom{52}{13}}[/tex]
