Correct Answer: C. \(\frac{5}{36}\)

Explanation

probability of selecting three balls with alternating colours = probability of selecting a blue ball (without replacement) x probability of selecting a red ball (without replacement) x probability of selecting a blue ball (without replacement)

In this case, we will be using the first instance (both would give the same answer since the number of blue and red balls are equal)

first: probability of selecting a red ball (without replacement) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

second: the probability of selecting a blue ball (without replacement) = \(\frac{5}{9}\)

third: probability of selecting a red ball (without replacement) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Therefore, the probability of selecting three balls with alternating colours = \(\frac{1}{2}\) x \(\frac{5}{9}\) x \(\frac{1}{2}\)

= \(\frac{5}{36}\)

Correct Answer: D. N5.25

Correct Answer: C. 60°

Correct Answer: B. \(10\frac{2}{3}cm^3\)

Correct Answer: A. 3/4

Correct Answer: D. \(- \cos x\)

Correct Answer: C. -4 \(\frac{7}{12}\)

Explanation

1\(\frac{3}{4} - (2 \frac{1}{3} + 4)\) = \(\frac{7}{4} - (\frac{7}{3} + \frac{4}{1})\); \(\frac{7 + 12}{3}\)

\(\frac{7}{4} - \frac{19}{3} = \frac{21 + 76}{12}\)

= \(\frac{-55}{12} = -4 \frac{7}{12}\)

Explanation

\(p(Ago) = \frac{4}{5} ; p(Sulley) = \frac{3}{4} ; p(Musa) = \frac{2}{3}\)

(a) p(none admitted) = \(p(Ago') \times p(Sulley') \times p(Musa')\)

= \(\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3}\)

= \(\frac{1}{60}\)

(b) p(Ago and Sulley admitted only) = \(p(Ago) \times p(Sulley) \times p(Musa')\)

= \(\frac{4}{5} \times \frac{3}{4} \times \frac{1}{3}\)

= \(\frac{1}{5}\)

Explanation

(a) Tie Edem and his wife. There are six people to arrange in 6! ways, Edem and his wife can be arranged in 2! ways.

\(\therefore\) Total number of ways = 2!6! = 1440 ways.

(b) 4 red, 5 black balls

Total = 9 balls.

\(p(red) = p = \frac{4}{9} ; p(black) = q = \frac{5}{9}\)

5 balls are selected. The binomial probability distribution function is

\((p + q)^{5} = p^{5} + 5p^{4}q + 10p^{3}q^{2} + 10p^{2}q^{3} + 5pq^{4} + q^{5}\)

(i) \(p(\text{red ball picked 3 times}) = 10p^{3}q^{2}\)

= \(10 \times (\frac{4}{9})^{3} \times (\frac{5}{9})^{2}\)

= \(10 \times \frac{64}{729} \times \frac{25}{81}\)

= \(0.271\)

(ii) \(p(\text{black ball at most 2 times}) = p(none) + p(once) + p(twice)\)

= \(10p^{3}q^{2} + 5p^{4}q + p^{5}\)

= \(\frac{16000}{59049} \times 5(\frac{4}{9})^{4} (\frac{5}{9}) + (\frac{4}{9})^{5}\)

= \(\frac{16000 + 6400 + 1024}{59049}\)

= \(\frac{23424}{59049}\)

= \(0.3967\)

Explanation

(a)

(c)(i) Position of lower quartile, \(Q_{1} = \frac{\sum f + 1}{4} = \frac{500 + 1}{4}\)

= \(\frac{501}{4} = 125.25th\) position.

\(\therefore Q_{1} = 26.0\)

Position of upper quartile, \(Q_{3} = 3(\frac{\sum f + 1}{4}) = 3(125.25) = 375.75th\) position.

\(\therefore Q_{3} = 56.3\)

Semi-interquartile range = \(\frac{Q_{3} - Q_{1}}{2} = \frac{56.3 - 26}{2}\)

= \(\frac{30.3}{2} = 15.15\)

(ii) If the pass mark is 37, then 190 students failed.

(iii) Those who scored between 0% and 20% = 100

Those who scored between 0% and 60% = 400

Those who scored between 20% and 60% = 400 - 100 = 300.

P(20% < x < 60%) = \(\frac{300}{500} = \frac{3}{5}\)