Dari seperangkat kartu bridge (remi) akan diambil satu kartu secara acak.tentukan peluang terambil satu kartu bukan As?

Matematika - Probabilitas (Peluang)

Pembahasan :

E = {(2♧), (3♧), (4♧), (5♧), (6♧), (7♧), (8♧), (9♧), (10♧), (J♧), (Q♧), (K♧), (2♢), (3♢), (4♢), (5♢), (6♢), (7♢), (8♢), (9♢), (10♢), (J♢), (Q♢), (K♢), (2♡), (3♡), (4♡), (5♡), (6♡), (7♡), (8♡), (9♡), (10♡), (J♡), (Q♡), (K♡), (2♤), (3♤), (4♤), (5♤), (6♤), (7♤), (8♤), (9♤), (10♤), (J♤), (Q♤), (K♤)}

n(E) = 48

S = {(A♧), (2♧), (3♧), (4♧), (5♧), (6♧), (7♧), (8♧), (9♧), (10♧), (J♧), (Q♧), (K♧), (A♢), (2♢), (3♢), (4♢), (5♢), (6♢), (7♢), (8♢), (9♢), (10♢), (J♢), (Q♢), (K♢), (A♡), (2♡), (3♡), (4♡), (5♡), (6♡), (7♡), (8♡), (9♡), (10♡), (J♡), (Q♡), (K♡), (A♤), (2♤), (3♤), (4♤), (5♤), (6♤), (7♤), (8♤), (9♤), (10♤), (J♤), (Q♤), (K♤)}

n(S) = 52

P(E) = n(E)/n(S)
= 48/52
= 12/13

PELUANG

1 set kartu bridge terdiri dari 4 gambar ♠️♥️♦️♣️), masing-masing diberi nomor/gambar 1 (as), 2, 3, ... 10, J, Q, K.
Seluruhnya terdiri dari 52 kartu.

terdapat 4 kartu as
dan (52-4) = 48 kartu bukan as.


PERCOBAAN :
Mengambil satu kartu dari 1 set kartu bridge sekali
n(S) = 52

KEJADIAN
A = kejadian terambil yang 1 kartu as
n(A) = 4

PELUANG
peluang terambilnya 1 kartu as
p(A) = n(A) / n(S)
p(A) = 4/52 = 1/13

peluang terambilnya kartu BUKAN kartu as
p(A') = 1 - p(A)
p(A') = 1 - 1/13 = 12/13 ✓


atau
KEJADIAN
B = kejadian terambilnya kartu bukan as
n(B) = 48

PELUANG
peluang terambilnya kartu BUKAN kartu as
p(B) = n(B) / n(S)
p(B) = 48/52 = 12/13 ✓