Respuesta :
Answer:
a
[tex]P(A_1) = 0.007[/tex]
b
[tex]P(A_2) = 0.016[/tex]
c
[tex]P(A_3) = 0.008[/tex]
d
[tex]P(A_4) = 0.004[/tex]
e
[tex]P(A_5) = 0.0002[/tex]
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
[tex]G = ^{13}C_5[/tex]
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
[tex]H = ^{39}C_2[/tex]
Generally the number of ways to select the 7 cards from the 52 deck of cards is
[tex]V = ^{52}C_7[/tex]
Generally the probability that exactly 5 of the 7 cards are hearts is mathematically represented as
[tex]P(A_1) = \frac{G * H}{V}[/tex]
=> [tex]P(A_1) = \frac{^{13}C_5 * ^{39}C_2}{^{52}C_7}[/tex]
=> [tex]P(A_1) = 0.007[/tex]
Considering question b
The number of ways to selected 3 diamonds out of the 13 diamonds is mathematically represented as
[tex] B = ^{13} C_3 [/tex]
The number of ways to selected 3 hearts out of the 13 hearts is mathematically represented as
[tex]K = ^{13}C_3[/tex]
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
[tex]M = ^{26}C_1[/tex]
Generally the probability that there are 3 hearts and 3 diamonds is
[tex]P(A_2) = \frac{B * K * M}{V}[/tex]
=> [tex]P(A_2) = \frac{^{13}C_3 * ^{13}C_3 * ^{26}C_1}{^{52}C_7}[/tex]
=> [tex]P(A_2) = 0.016[/tex]
Considering question c
The number of ways to selected 1 spade out of the 13 spade is mathematically represented as
[tex]J = ^{13}C_1[/tex]
Generally the probability that there are 3 hearts, 3 diamonds and 1 spade is
[tex]P(A_3) = \frac{B * K * J}{V}[/tex]
=> [tex]P(A_3) = \frac{^{13}C_3 * ^{13}C_3 * ^{13}C_1}{^{52}C_7}[/tex]
=> [tex]P(A_3) = 0.008[/tex]
Considering question d
The number of ways to selected 2 Aces out of the 4 Aces is mathematically represented a
[tex]U = ^{4}C_2[/tex]
The number of ways to selected 2 Kings out of the 4 Kings is mathematically represented a
[tex]R = ^{4}C_2[/tex]
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
[tex]S = ^{44}C_3[/tex]
Generally the probability that there are 2 Aces and 2 Kings is
[tex]P(A_4) = \frac{U * R * S}{ V}[/tex]
=> [tex]P(A_4) = \frac{ ^{4}C_2 * ^{4}C_2 * ^{44}C_3}{ ^{52}C_7}[/tex]
=>[tex]P(A_4) = 0.004[/tex]
Considering question e
The number of ways to selected 3 Kings out of the 4 Kings is mathematically represented a
[tex]P = ^{4}C_3[/tex]
Generally the number of ways of selecting the reaming 2 cards is
[tex]Q = ^{44}C_2[/tex]
Generally the probability that there are 2 Aces and 3 Kings is
[tex]P(A_5) = \frac{U * P * Q }{V}[/tex]
=> [tex]P(A_5) = \frac{ ^{4}C_2 * ^{4}C_3 * ^{44}C_2 }{^{52}C_7}[/tex]
=> [tex]P(A_5) = 0.0002[/tex]