5.3.1 Taburan Kebarangkalian, SPM Praktis (Kertas 1)

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} \frac{1}{16} + \frac{1}{4} + h + \frac{1}{4} + \frac{1}{16} = 1 \\ h = 1 - \frac{5}{8} \\ h = \frac{3}{8} \end{array}

(b)

P (X ≥ 3) = P (X = 3) + P (X = 4)

P (X \geq 3) = \frac{1}{4} + \frac{1}{16} = \frac{5}{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} Sisihan piawai = \sqrt{n p q} \\ = \sqrt{10 \times \frac{1}{4} \times \frac{3}{4}} \\ = 1.875 \end{array}

(b)

\begin{array}{l} P (X = r) =^{10} C_{r} {(\frac{1}{4})}^{r} {(\frac{3}{4})}^{10 - r} \\ P (X \geq 1) \\ = 1 - P (X < 1) \\ = 1 - P (X = 0) \\ = 1 -^{10} C_{0} {(\frac{1}{4})}^{0} {(\frac{3}{4})}^{10} \\ = 0.9437 \end{array}