Question

A flush in the card game of poker occurs if a player gets five cards that are all the same suit​ (clubs, diamonds,​ hearts, or​ spades). Complete parts​ (a) and​ (b) to obtain the probability of being dealt a flush in five cards. ​(a) Initially concentrate on one​ suit, say spades. There are 13 spades in a deck. Compute​ P(five spades​)equals​P(first card is spades and second card is spades and third card is spades and fourth card is spades and fifth card is spades​). ​P(five spades​)equals nothing

Answer 1

Answer:

The probability is 0.000495

Step-by-step explanation:

As per the question:

Total no. of cards in a deck = 52

No. of spades in a deck = 13

Now, we have to select 5 cards in a deck such that they belong to the same suit, i.e., spades.

The no. of ways of selecting 5 cards from a deck = $^{52}C_{5} = \frac{52!}{5!(52 - 5)!} = \frac{52!}{5!47!}$

The no. of ways of selecting 5 cards from 13 spade cards = $^{13}C_{5} = \frac{13!}{5!(13 - 5)!} = \frac{13!}{5!8!}$

Now,

Probability that the selected 5 cards are all spades, P(E)  = $\frac{No.\ of\ ways\ of\ selecting\ 5\ cards\ from\ 13\ spade\ cards}{No.\ of\ ways\ of\ selecting\ 5\ cards\ from\ a\ deck}$

P(E)  = $\frac{^{13}C_{5}}{^{52}C_{5}}$

P(E)  = $\frac{\frac{13!}{5!8!}}{\frac{52!}{5!52!}}$

P(E)  = $\frac{\frac{13\times 12\times 11\tiems 10\times 9}{5\times 4\times 3\times 2\times 1}}{\frac{52\times 51\times 50\times 49\times 48}{5\times 4\times 3\times 2\times 1}}$

P(E)  = $\frac{13\times 12\times 11\tiems 10\times 9}{52\times 51\times 50\times 49\times 48} = 4.95\times 10^{- 4} = 0.000495$