2021-I-15

26-06-2023 app.cch No Comments

Ans: (a)

3 628 800

(b)

\frac{7}{15}

The required number of queues
$\begin{array}{cl} = & (7 + 3)! \\ = & 3 628 800 \end{array}$
The required probability
$\begin{array}{cl} = & \frac{7! \times P_{3}^{8}}{3 628 800} \\ = & \frac{7}{15} \end{array}$

2021, HKDSE-MATH, Paper 1 Probability