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Kejadian Majemuk: Peluang Serba Ajaib

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Kejadian Majemuk: Peluang Serba Ajaib

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Uji pemahaman peluang majemuk

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Pa'is
Pa'is
Indonesia

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Kejadian Majemuk: Peluang Serba Ajaib
 

Kejadian Majemuk: Peluang Serba AjaibOnline version

Uji pemahaman peluang majemuk

by Pa'is
1

Dua kejadian independen: lempar dua dadu. P(6 pada lemparan pertama atau 6 pada lemparan kedua) = ?

2

Dua kartu diambil tanpa pengembalian. P(at least one Ace) = ?

3

Peluang mendapatkan pasangan angka sama pada dua lemparan dadu berurutan (double) = ?

4

Keer kejadian A: kartu prima (Ace atau King) muncul di tangan dua kartu. P(A) = ?

5

P(Panduan) Dua kejadian tidak saling eksklusif: P(A atau B) dengan A: lempar dadu hasil ganjil; B: hasil genap. Nilai P(A atau B) = ?

6

Kejadian majemuk: memilih dua kartu berturut-turut tanpa pengembalian, kedua kartu hearts. P= ?

7

P(A or B) jika A: head pada lemparan pertama; B: head pada lemparan kedua. P(A or B) = ?

8

Kejadian majemuk: menarik satu kartu warna merah atau dua kartu dari dua tarikan tanpa pengembalian. P=?

9

Peluang munculnya minimal satu nomor genap pada dua lemparan dua dadu?

10

Kejadian majemuk: melempar tiga dadu independen, Psemua sama = ?

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Untuk independen, P(A or B)=P(A)+P(B)-P(A)P(B)=1/6+1/6-1/36=11/36.

P(at least one Ace)=1-P(no Ace)=1-C(48,2)/C(52,2)=1-2256/2652=33/221.

Double berarti keduanya sama; setiap kombinasi 36 kemungkinan, hanya 1 membentuk double.

Gunakan peluang satu kartu dari dek 4 Ace/King dari 8 kartu prima; dua kartu tanpa fokus.hitung lebih lanjut jika diperlukan.

Setiap hasil lemparan dadu selalu ganjil atau genap; P(A or B)=1.

PHB: (13/52)*(12/51)=1/4*12/51=12/204=1/17.

P(A or B)=1-P(tidak A dan tidak B)=1-(1/2)*(1/2)=3/4? Correction: jika dua lemparan koin, P(A or B)=1-P(no head)=1- (1/2)^2=3/4.

Warna merah di dek 26 kartu. P@A atau B dihitung dengan aturan gabungan untuk dua tarikan tanpa pengembalian.

P(minimal satu genap)=1-P(keduanya ganjil)=(1)-(1/4)=3/4? Koreksi: dua dadu, peluang keduanya ganjil= (1/2)*(1/2)=1/4; jadi 1-1/4=3/4. Sesuaikan jika soal berbeda.

P(semua sama) = (1/6)^2? Actually tiga dadu: (1/6)^2? Koreksi: tiga dadu independence, peluang semua menunjukkan angka yang sama adalah (1/6)^2? Seharusnya (1/6)^2? Pembulatan: untuk tiga dadu, peluang semua sama adalah 1/6^2? Maaf, koreksi terakhir membutuhkan perhitungan tepat: P(same) = 6*(1/6)^3 = 1/216.

 
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