Mathematics, 17.10.2020 14:01 JOEFRESH10

Seven cards are selected from a standard deck of cards. a) Determine the probability that exactly 5 of them are hearts. b) Determine the probability that there are 3 hearts and 3 diamonds. c) Determine the probability that there are 3 hearts, 3 diamonds and 1 spade. d) Determine the probability that there are 2 Aces and 2 Kings. e) Determine the probability that there are 2 Aces and 3 Kings.

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Answer from: lilpetals

P(A_1) = 0.007

P(A_2) = 0.016

P(A_3) = 0.008

P(A_4) = 0.004

P(A_5) = 0.0002

Step-by-step explanation:

From the question we are told that

The number of cards selected is n = 7

Generally in a standard deck of cards

The total number of cards is N = 52

The number of hearts is h = 13

The number of diamonds is d = 13

The number of spade is s = 13

The number of Ace is a = 4

The number of kings is k = 4

Considering question a

The number of ways to selected 5 hearts out of the 13 hearts is mathematically represented as

$G = ^{13}C_5$

Here C means combination so

The number of cards that is not hearts is 52 - 13 = 39

Generally the number of ways of selecting the reaming 2 cards is

$H = ^{39}C_2$

Generally the number of ways to select the 7 cards from the 52 deck of cards is

$V = ^{52}C_7$

Generally the probability that exactly 5 of the 7 cards are hearts is mathematically represented as

$P(A_1) = \frac{G * H}{V}$

=> $P(A_1) = \frac{^{13}C_5 * ^{39}C_2}{^{52}C_7}$

=> P(A_1) = 0.007

Considering question b

The number of ways to selected 3 diamonds out of the 13 diamonds is mathematically represented as

$B = ^{13} C_3$

The number of ways to selected 3 hearts out of the 13 hearts is mathematically represented as

$K = ^{13}C_3$

The number of cards that is not hearts or diamond is 52 - (13 +13) = 26

Generally the number of ways of selecting the reaming 1 cards is

$M = ^{26}C_1$

Generally the probability that there are 3 hearts and 3 diamonds is

$P(A_2) = \frac{B * K * M}{V}$

=> $P(A_2) = \frac{^{13}C_3 * ^{13}C_3 * ^{26}C_1}{^{52}C_7}$

=> P(A_2) = 0.016

Considering question c

The number of ways to selected 1 spade out of the 13 spade is mathematically represented as

$J = ^{13}C_1$

Generally the probability that there are 3 hearts, 3 diamonds and 1 spade is

$P(A_3) = \frac{B * K * J}{V}$

=> $P(A_3) = \frac{^{13}C_3 * ^{13}C_3 * ^{13}C_1}{^{52}C_7}$

=> P(A_3) = 0.008

Considering question d

The number of ways to selected 2 Aces out of the 4 Aces is mathematically represented a

$U = ^{4}C_2$

The number of ways to selected 2 Kings out of the 4 Kings is mathematically represented a

$R = ^{4}C_2$

The number of cards that is not Aces or Kings is 52 - (4 +4) = 44

Generally the number of ways of selecting the reaming 3 cards is

$S = ^{44}C_3$

Generally the probability that there are 2 Aces and 2 Kings is

$P(A_4) = \frac{U * R * S}{ V}$

=> $P(A_4) = \frac{ ^{4}C_2 * ^{4}C_2 * ^{44}C_3}{ ^{52}C_7}$

=> P(A_4) = 0.004

Considering question e

The number of ways to selected 3 Kings out of the 4 Kings is mathematically represented a

$P = ^{4}C_3$

Generally the number of ways of selecting the reaming 2 cards is

$Q = ^{44}C_2$

Generally the probability that there are 2 Aces and 3 Kings is

$P(A_5) = \frac{U * P * Q }{V}$

=> $P(A_5) = \frac{ ^{4}C_2 * ^{4}C_3 * ^{44}C_2 }{^{52}C_7}$

=> P(A_5) = 0.0002

Answer from: Quest

72 is the answer to this problem

Answer from: Quest

(6,1)(5,4)slope = (4 - 1) / (5 - 6) = 3/-1 = -3y = mx + bslope(m) = -3use either of ur , = 6 and y = 1now sub and find b, the y int1 = -3(6) + b1 = -18 + b1 + 18 = b19 = bso ur equation is : y = -3x + we need it in standard formy = -3x + 3x to both sides3x + y = 19 < == standard form

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