Tentukan probabilitas menerka secara tepat paling sedikit 7 dari 10 soal ujian tipe benar-salah.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah 0 , 172 . Ingat P ( X , n ) = C ( n , x ) ⋅ p x ⋅ q n − x Keterangan: P ( X , n ) = probabilitas kejadian X dalam n percobaan X = kejadian sukses C = kombinasi p = probabilitas sukses q = probabilitas gagal n = banyak percobaan x = banyak kejadian sukses Tipe soal benar atau salah, misal diasumsikan kejadian sukses adalah " pilihan benar" maka peluang sukses sebagai berikut. p = 2 1 ​ , berarti q = 1 − p = 2 1 ​ Jumlah soal ada 10 , artinya n = 10 maka probabilitas kejadian menerka paling sedikit 7 soal sebagai berikut. P ( X ​ ≥ ​ 7 ) = P ( X = 7 ) + P ( X = 8 ) + P ( X = 9 ) + P ( X = 10 ) ​ P ( X ​ = ​ 7 ) ​ ​ = ​ ​ P ( x ​ ​ = ​ ​ 7 , n ​ = ​ 10 ) ​ = ​ ​ C ( 10 , 7 ) ⋅ ( 2 1 ​ ) 7 ⋅ ( 2 1 ​ ) 10 − 7 ​ = 7 ! 3 ! 10 ! ​ ​ ​ ⋅ ​ ​ ​ ( 2 1 ​ ) ​ 7 ​ ​ ⋅ ​ ​ ​ ( 2 1 ​ ) ​ 3 = 2 10 120 ​ ​ P ( X ​ = ​ 8 ) ​ ​ = ​ ​ P ( x ​ ​ = ​ ​ 8 , n ​ = ​ 10 ) ​ = ​ ​ C ( 10 , 8 ) ⋅ ( 2 1 ​ ) 8 ⋅ ( 2 1 ​ ) 10 − 8 ​ = 8 ! 2 ! 10 ! ​ ​ ​ ⋅ ​ ​ ​ ( 2 1 ​ ) ​ 8 ⋅ ​ ​ ( 2 1 ​ ) 2 ​ = 2 10 45 ​ ​ P ( X ​ = ​ 9 ) ​ ​ = ​ ​ P ( x ​ ​ = ​ ​ 9 , n ​ = ​ 10 ) ​ = ​ ​ C ( 10 , 9 ) ⋅ ( 2 1 ​ ) 8 ⋅ ( 2 1 ​ ) 10 − 9 ​ = 9 ! 1 ! 10 ! ​ ​ ​ ⋅ ​ ​ ​ ( 2 1 ​ ) ​ 9 ⋅ ​ ​ ( 2 1 ​ ) ​ = 2 10 10 ​ ​ P ( X ​ = ​ 8 ) ​ ​ = ​ ​ P ( x ​ ​ = ​ ​ 8 , n ​ = ​ 10 ) ​ = ​ ​ C ( 10 , 8 ) ⋅ ( 2 1 ​ ) 10 ⋅ ( 2 1 ​ ) 10 − 10 ​ = 10 ! 0 ! 10 ! ​ ​ ​ ⋅ ​ ​ ​ ( 2 1 ​ ) ​ 10 = 2 10 1 ​ ​ P ( X ≥ 7 ) ​ ​ = ​ P ( X = 7 ) + P ( X = 8 ) + P ( X = 9 ) + P ( X = 10 ) ​ = 2 10 120 ​ + 2 10 45 ​ + 2 10 10 ​ + 2 10 1 ​ = 2 10 176 ​ = 0 , 172 ​ Dengan demikian,probabilitas menerka secara tepat paling sedikit 7 dari 10 soal ujian tipe benar-salah adalah 0 , 172.