All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Firstly, the number of different friends is a fixed constant.
Secondly, the probability of successfully contacting a friend is constant.

(ii)
The probability of contacting a friend will vary from friend to friend, as some friend might not be free or contactable during that evening.

(iii)
$\mathrm{R} \sim \mathrm{B}(8,0.7)$

$\mathrm{P}(R \ge 6) = 1 - \mathrm{P}(R \le 5) \approx 0.552$

(iv)
$\mathrm{R} \sim \mathrm{B}(40, 0.7)$

Since $n = 40 \mathrm{~is~large,~} np = 28 > 5 \mathrm{~and~} nq = 12 > 5 ~ \mathrm{R} \sim \mathrm{N}(28, 8.4) \mathrm{approximately}$

$\mathrm{P}(R \textless 25)$

$= \mathrm{P}(R \textless 24.5)$ by continuity correction

$\approx 0.114$