5.3.1 Taburan Kebarangkalian, SPM Praktis (Kertas 1)
Soalan 1:
Rajah di bawah menunjukkan graf suatu taburan binomial bagi X.
Cari
(a) nilai h,
(b) P (X ≥ 3)
Penyelesaian:
(a)
P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1
\(\begin{array}{l}\frac1{16}+\frac14+h+\frac14+\frac1{16}=1\\h=1-\frac58\\h=\frac38\end{array}\)
(b)
P (X ≥ 3) = P (X = 3) + P (X = 4)
\(P{(X\geq3)}=\frac14+\frac1{16}=\frac5{16}\)
Soalan 2:
Pembolehubah rawak X mewakili taburan binomial dengan 10 percubaan dan keberangkalian berjaya ialah ¼.
(a) sisihan piawai taburan itu,
(b) kebarangkalian bahawa sekurang-kurangnya satu percubaan adalah berjaya.
Penyelesaian:
(a)
n = 10, p = ¼
\(\begin{array}{l}\text{Sisihan piawai}=\sqrt{npq}\\=\sqrt{10\times\frac14\times\frac34}\\=1.875\end{array}\)
(b)
\(\begin{array}{l}P(X=r)={}^{10}C_r{(\frac14)}^r{(\frac34)}^{10-r}\\P(X\geq1)\\=1-P(X<1)\\=1-P{(X=0)}\\=1-{}^{10}C_0{(\frac14)}^0{(\frac34)}^{10}\\=0.9437\end{array}\)