Pertanyaan

David mendapat tugas mengerjakan 10 soal pilihan ganda 5 opsi jawaban A, B, C, D dan E. David hanya menebak jawaban setiap soal. Tentukan peluang David menjawab benar paling banyak 9 soal!

David mendapat tugas mengerjakan $10$ soal pilihan ganda $5$ opsi jawaban A, B, C, D dan E. David hanya menebak jawaban setiap soal. Tentukan peluang David menjawab benar paling banyak $9$ soal!

S. Dwi

Master Teacher

Jawaban terverifikasi

Jawaban

peluang David menjawab benar paling banyak 9 soal adalah 0 , 9999 .

peluang David menjawab benar paling banyak

9

soal adalah

0, 9999

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah 0 , 9999 . Fungsi Distribusi Binomial Kumulatif Peluang paling banyak x kejadian yang diharapkan dinamakan fungsi distribusi binomial kumulatif. Misalkan x = t , maka peluang paling banyak t kejadian yang diharapkan dinyatakan dengan: f ( t ) = P ( X ≤ t ) = x = 0 ∑ t C ( n , x ) ⋅ p x ⋅ q n − x Berdasarkan soal di atas diketahui n = 10 , p = 5 1 , q = 5 4 dan t = 9 . Dengan menggunakan rumus di atas, dapat kita tentukan P ( X ≤ 9 ) sebagai berikut: Untuk P ( X = 0 ) , maka: P ( X = 0 ) = = = = C ( 10 , 0 ) ⋅ ( 5 1 ) 0 ⋅ ( 5 4 ) 10 − 0 0 ! ( 10 − 0 ) ! 10 ! ⋅ ( 1 ) ⋅ ( 5 10 4 10 ) ( 1 ) ⋅ ( 9.765.625 1.048.576 ) 0 , 1074 Untuk P ( X = 1 ) , maka: P ( X = 1 ) = = = = = = C ( 10 , 1 ) ⋅ ( 5 1 ) 1 ⋅ ( 5 4 ) 10 − 1 1 ! ( 10 − 1 ) ! 10 ! ⋅ ( 5 1 ) ⋅ ( 5 9 4 9 ) 1 ( 9 ! ) 10 ⋅ 9 ! ⋅ ( 5 1 ) ⋅ ( 1.953.125 262.144 ) 10 ⋅ ( 9.765.625 262.144 ) 9.765.625 2.621.440 0 , 2684 Untuk P ( X = 2 ) maka: P ( X = 2 ) = = = = = = C ( 10 , 2 ) ⋅ ( 5 1 ) 2 ⋅ ( 5 4 ) 10 − 2 2 ! ( 10 − 2 ) ! 10 ! ⋅ ( 5 2 1 2 ) ⋅ ( 5 8 4 8 ) ( 2 ⋅ 1 ) ( 8 ! ) 10 ⋅ 9 ⋅ 8 ! ⋅ ( 25 1 ) ⋅ ( 390.625 65.536 ) 45 ⋅ ( 9.765.625 65.536 ) 9.765.625 2.949.120 0 , 302 Untuk P ( X = 3 ) maka: P ( X = 3 ) = = = = = = = C ( 10 , 3 ) ⋅ ( 5 1 ) 3 ⋅ ( 5 4 ) 10 − 3 3 ! ( 10 − 3 ) ! 10 ! ⋅ ( 125 1 ) ⋅ ( 5 7 4 7 ) ( 3 ⋅ 2 ⋅ 1 ) ( 7 ! ) 10 ⋅ 9 ⋅ 8 ⋅ 7 ! ⋅ ( 125 1 ) ⋅ ( 78.125 16.384 ) ( 10 ⋅ 3 ⋅ 4 ) ⋅ ( 9.765.625 16.384 ) ( 120 ) ⋅ ( 9.765.625 16.384 ) 9.765.625 1.966.080 0 , 2013 Untuk P ( X = 4 ) maka: P ( X = 4 ) = = = = = = = C ( 10 , 4 ) ⋅ ( 5 1 ) 4 ⋅ ( 5 4 ) 10 − 4 4 ! ( 10 − 4 ) ! 10 ! ⋅ ( 625 1 ) ⋅ ( 5 6 4 6 ) ( 4 ⋅ 3 ⋅ 2 ⋅ 1 ) ( 6 ! ) 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ! ⋅ ( 625 1 ) ⋅ ( 15.625 4.096 ) ( 10 ⋅ 3 ⋅ 7 ) ⋅ ( 9.765.625 4.096 ) ( 210 ) ⋅ ( 9.765.625 4.096 ) 9.765.625 860.160 0 , 0881 Untuk P ( X = 5 ) maka: P ( X = 5 ) = = = = = = = C ( 10 , 5 ) ⋅ ( 5 1 ) 5 ⋅ ( 5 4 ) 10 − 5 5 ! ( 10 − 5 ) ! 10 ! ⋅ ( 3.125 1 ) ⋅ ( 5 5 4 5 ) ( 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ) ( 5 ! ) 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ! ⋅ ( 9.765.625 1.024 ) ( 2 ⋅ 9 ⋅ 2 ⋅ 7 ) ⋅ ( 9.765.625 1.024 ) ( 252 ) ⋅ ( 9.765.625 1.024 ) 9.765.625 258.048 0 , 0264 Untuk P ( X = 6 ) maka: P ( X = 6 ) = = = = = = = C ( 10 , 6 ) ⋅ ( 5 1 ) 6 ⋅ ( 5 4 ) 10 − 6 6 ! ( 10 − 6 ) ! 10 ! ⋅ ( 15.625 1 ) ⋅ ( 5 4 4 4 ) 6 ! ( 4 ⋅ 3 ⋅ 2 ⋅ 1 ) 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ! ⋅ ( 15.625 1 ) ⋅ ( 625 256 ) ( 10 ⋅ 3 ⋅ 7 ) ⋅ ( 9.765.625 256 ) ( 210 ) ⋅ ( 9.765.625 256 ) 9.765.625 53.760 0 , 0055 Untuk P ( X = 7 ) maka: P ( X = 7 ) = = = = = = = C ( 10 , 7 ) ⋅ ( 5 1 ) 7 ⋅ ( 5 4 ) 10 − 7 7 ! ( 10 − 7 ) ! 10 ! ⋅ ( 15.625 1 ) ⋅ ( 5 3 4 3 ) 7 ! ( 3 ⋅ 2 ⋅ 1 ) 10 ⋅ 9 ⋅ 8 ⋅ 7 ! ⋅ ( 78.125 1 ) ⋅ ( 125 64 ) ( 10 ⋅ 3 ⋅ 4 ) ⋅ ( 9.765.625 64 ) ( 120 ) ⋅ ( 9.765.625 64 ) 9.765.625 7.680 0 , 0008 Untuk P ( X = 8 ) maka: P ( X = 8 ) = = = = = = C ( 10 , 8 ) ⋅ ( 5 1 ) 8 ⋅ ( 5 4 ) 10 − 8 8 ! ( 10 − 8 ) ! 10 ! ⋅ ( 390.625 1 ) ⋅ ( 5 2 4 2 ) 8 ! ( 2 ! ) 10 ⋅ 9 ⋅ 8 ! ( 9.765.625 16 ) ( 45 ) ( 9.765.625 16 ) 9.765.625 720 0 , 00007 Untuk P ( X = 9 ) maka: P ( X = 9 ) = = = = = = C ( 10 , 9 ) ⋅ ( 5 1 ) 9 ⋅ ( 5 4 ) 10 − 9 9 ! ( 10 − 9 ) ! 10 ! ⋅ ( 1.953 , 125 1 ) ⋅ ( 5 4 ) 9 ! 1 ! 10 ⋅ 9 ! ( 9.765.625 4 ) ( 10 ) ( 9.765.625 4 ) 9.765.625 40 0 , 000004 Diperoleh: P ( X ≤ 9 ) = = ∑ x = 0 9 C ( 10 , 9 ) ⋅ ( 5 1 ) 9 ⋅ ( 5 4 ) 1 0 , 9999 Dengan demikian, peluang David menjawab benar paling banyak 9 soal adalah 0 , 9999 .

Jawaban yang benar untuk pertanyaan tersebut adalah $0, 9999$ .

Fungsi Distribusi Binomial Kumulatif

Peluang paling banyak $x$ kejadian yang diharapkan dinamakan fungsi distribusi binomial kumulatif. Misalkan $x = t$ , maka peluang paling banyak $t$ kejadian yang diharapkan dinyatakan dengan:

$f (t) = P (X \leq t) = x = 0 \sum t C (n, x) \cdot p^{x} \cdot q^{n - x}$

Berdasarkan soal di atas diketahui $n = 10$ , $p = \frac{1}{5}$ , $q = \frac{4}{5}$ dan $t = 9$ . Dengan menggunakan rumus di atas, dapat kita tentukan $P (X \leq 9)$ sebagai berikut:

Untuk $P (X = 0)$ , maka:

$P (X = 0) = = = = C (10, 0) \cdot (\frac{1}{5})^{0} \cdot (\frac{4}{5})^{10 - 0} \frac{10 !}{0 ! ( 10 - 0 ) !} \cdot (1) \cdot (\frac{4 ^{10}}{5 ^{10}}) (1) \cdot (\frac{1.048.576}{9.765.625}) 0, 1074$

Untuk $P (X = 1)$ , maka:

$P (X = 1) = = = = = = C (10, 1) \cdot (\frac{1}{5})^{1} \cdot (\frac{4}{5})^{10 - 1} \frac{10 !}{1 ! ( 10 - 1 ) !} \cdot (\frac{1}{5}) \cdot (\frac{4 ^{9}}{5 ^{9}}) \frac{10 \cdot 9 !}{1 ( 9 ! )} \cdot (\frac{1}{5}) \cdot (\frac{262.144}{1.953.125}) 10 \cdot (\frac{262.144}{9.765.625}) \frac{2.621.440}{9.765.625} 0, 2684$

Untuk $P (X = 2)$ maka:

$P (X = 2) = = = = = = C (10, 2) \cdot (\frac{1}{5})^{2} \cdot (\frac{4}{5})^{10 - 2} \frac{10 !}{2 ! ( 10 - 2 ) !} \cdot (\frac{1 ^{2}}{5 ^{2}}) \cdot (\frac{4 ^{8}}{5 ^{8}}) \frac{10 \cdot 9 \cdot 8 !}{( 2 \cdot 1 ) ( 8 ! )} \cdot (\frac{1}{25}) \cdot (\frac{65.536}{390.625}) 45 \cdot (\frac{65.536}{9.765.625}) \frac{2.949.120}{9.765.625} 0, 302$

Untuk $P (X = 3)$ maka:

$P (X = 3) = = = = = = = C (10, 3) \cdot (\frac{1}{5})^{3} \cdot (\frac{4}{5})^{10 - 3} \frac{10 !}{3 ! ( 10 - 3 ) !} \cdot (\frac{1}{125}) \cdot (\frac{4 ^{7}}{5 ^{7}}) \frac{10 \cdot 9 \cdot 8 \cdot 7 !}{( 3 \cdot 2 \cdot 1 ) ( 7 ! )} \cdot (\frac{1}{125}) \cdot (\frac{16.384}{78.125}) (10 \cdot 3 \cdot 4) \cdot (\frac{16.384}{9.765.625}) (120) \cdot (\frac{16.384}{9.765.625}) \frac{1.966.080}{9.765.625} 0, 2013$

Untuk $P (X = 4)$ maka:

$P (X = 4) = = = = = = = C (10, 4) \cdot (\frac{1}{5})^{4} \cdot (\frac{4}{5})^{10 - 4} \frac{10 !}{4 ! ( 10 - 4 ) !} \cdot (\frac{1}{625}) \cdot (\frac{4 ^{6}}{5 ^{6}}) \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 !}{( 4 \cdot 3 \cdot 2 \cdot 1 ) ( 6 ! )} \cdot (\frac{1}{625}) \cdot (\frac{4.096}{15.625}) (10 \cdot 3 \cdot 7) \cdot (\frac{4.096}{9.765.625}) (210) \cdot (\frac{4.096}{9.765.625}) \frac{860.160}{9.765.625} 0, 0881$

Untuk $P (X = 5)$ maka:

$P (X = 5) = = = = = = = C (10, 5) \cdot (\frac{1}{5})^{5} \cdot (\frac{4}{5})^{10 - 5} \frac{10 !}{5 ! ( 10 - 5 ) !} \cdot (\frac{1}{3.125}) \cdot (\frac{4 ^{5}}{5 ^{5}}) \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 !}{( 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 ) ( 5 ! )} \cdot (\frac{1.024}{9.765.625}) (2 \cdot 9 \cdot 2 \cdot 7) \cdot (\frac{1.024}{9.765.625}) (252) \cdot (\frac{1.024}{9.765.625}) \frac{258.048}{9.765.625} 0, 0264$

Untuk $P (X = 6)$ maka:

$P (X = 6) = = = = = = = C (10, 6) \cdot (\frac{1}{5})^{6} \cdot (\frac{4}{5})^{10 - 6} \frac{10 !}{6 ! ( 10 - 6 ) !} \cdot (\frac{1}{15.625}) \cdot (\frac{4 ^{4}}{5 ^{4}}) \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 !}{6 ! ( 4 \cdot 3 \cdot 2 \cdot 1 )} \cdot (\frac{1}{15.625}) \cdot (\frac{256}{625}) (10 \cdot 3 \cdot 7) \cdot (\frac{256}{9.765.625}) (210) \cdot (\frac{256}{9.765.625}) \frac{53.760}{9.765.625} 0, 0055$

Untuk $P (X = 7)$ maka:

$P (X = 7) = = = = = = = C (10, 7) \cdot (\frac{1}{5})^{7} \cdot (\frac{4}{5})^{10 - 7} \frac{10 !}{7 ! ( 10 - 7 ) !} \cdot (\frac{1}{15.625}) \cdot (\frac{4 ^{3}}{5 ^{3}}) \frac{10 \cdot 9 \cdot 8 \cdot 7 !}{7 ! ( 3 \cdot 2 \cdot 1 )} \cdot (\frac{1}{78.125}) \cdot (\frac{64}{125}) (10 \cdot 3 \cdot 4) \cdot (\frac{64}{9.765.625}) (120) \cdot (\frac{64}{9.765.625}) \frac{7.680}{9.765.625} 0, 0008$

Untuk $P (X = 8)$ maka:

$P (X = 8) = = = = = = C (10, 8) \cdot (\frac{1}{5})^{8} \cdot (\frac{4}{5})^{10 - 8} \frac{10 !}{8 ! ( 10 - 8 ) !} \cdot (\frac{1}{390.625}) \cdot (\frac{4 ^{2}}{5 ^{2}}) \frac{10 \cdot 9 \cdot 8 !}{8 ! ( 2 ! )} (\frac{16}{9.765.625}) (45) (\frac{16}{9.765.625}) \frac{720}{9.765.625} 0, 00007$

Untuk $P (X = 9)$ maka:

$P (X = 9) = = = = = = C (10, 9) \cdot (\frac{1}{5})^{9} \cdot (\frac{4}{5})^{10 - 9} \frac{10 !}{9 ! ( 10 - 9 ) !} \cdot (\frac{1}{1.953 , 125}) \cdot (\frac{4}{5}) \frac{10 \cdot 9 !}{9 ! 1 !} (\frac{4}{9.765.625}) (10) (\frac{4}{9.765.625}) \frac{40}{9.765.625} 0, 000004$

Diperoleh:

$P (X \leq 9) = = \sum_{x = 0}^{9} C (10, 9) \cdot (\frac{1}{5})^{9} \cdot (\frac{4}{5})^{1} 0, 9999$

Dengan demikian, peluang David menjawab benar paling banyak $9$ soal adalah $0, 9999$ .

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