Question Library

The probability not scoring 85% is = 1 - pro(scoring 85%) = 1 - \(\frac{85}{100}\) = 1 - 0.85 = 0.15 not scoring would be less than 1. Hence option D is correct because the probability of an event cannot be more than 1.

\(\frac{x}{7} = \frac{96}{1}\) ==> \(\frac{672 + x}{8} = 112\) Therefore x = 224

Arrange all the values in ascending order, 2,2,3,3,3,4,4,4,4,5,5,5,7,8,9,9

Pass(P) = \(\frac{2}{3}\), Fail(F) = \(\frac{1}{3}\) T = P.P.F ==> \(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}\) = \(\frac{4}{27}\)

Man lives = \(\frac{2}{3}\) not live = \(\frac{1}{3}\) Wife lives = \(\frac{3}{5}\) not live = \(\frac{2}{5}\) P(at least one lives to 80 years) = P(man lives to 80 not woman) + P(woman lives to 80 and not man) + P(both live to 80) \(P = (\frac{2}{3} \times \frac{2}{5}) + (\frac{2}{5} \times \frac{1}{3}) + (\frac{2}{3} \times \frac{3}{5})\) = \(\frac{4}{15} + \frac{3}{15} + \frac{6}{15}\) = \(\frac{13}{15}\)

Prob(RB + BR) = Total balls = 4 + 6 = 10 = prob(\(\frac{4}{10} \times \frac{6}{9}\)) + prob(\(\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90}\) = \(\frac{48}{90} = \frac{16}{30} = \frac{8}{15}\)

6 + 8 + 6 + 5 = 25

\(\begin{array}{c|c} x & f & fx\\ \hline 1 & 6 & 6 \\ 2 & 8 & 16\\ 3 & 8 & 18\\ 4 & 5 & 20\end{array}\) mean x = \(\frac{\sum fx}{\sum f}\) = \(\frac{60}{25}\) x = 2.4

median = (\(\frac{N + 1}{2}\))th = (\(\frac{25 + 1}{2}\))th = \(\frac{26th}{2}\) = 13th the 13th is 2

(a) (i) No of red discs = \(\frac{1}{4} \times 40 = 10\) Total no of discs = 40 \(\therefore\) No of white balls = 40 - 10 = 30 (ii) Let the number of red balls to be added = z \(\therefore \frac{10 + z}{40 + z} = \frac{1}{3}\) \(3(10 + z) = 40 + z\) \(30 + 3z = 40 + z \implies 3z - z = 40 - 30\) \(2z = 10 \implies z = 5\) Therefore, adding 5 red discs, the probability of picking a red disc = \(\frac{1}{3}\). (b) CP of a plate = N50.00 Let the number of plates be x \(\therefore \text{CP of x plates} = N50x\) \(\text{SP of all plates} = N600\) \(\frac{100 + 20}{100} \times N50x = N600 \implies N60x = N600\) \(x = \frac{600}{60} = 10 plates\)

Mode = L1 + (\(\frac{D_1}{D_1 + D_2}\))C D1 = frequency of modal class - frequency of the class before it D1 = 5 - 2 = 3 D2 = frequency of modal class - frequency of the class that offers it D2 = 5 - 3 = 2 L1 = lower class boundary of the modal class L1 = 5 - 5 C is the class width = 8 - 5.5 = 3 Mode = L1 + (\(\frac{D_1}{D_1 + D_2}\))C = 5.5 + \(\frac{3}{2 + 3}\)C = 5.5 + \(\frac{3}{5}\) x 3 = 5.5 + \(\frac{9}{5}\) = 5.5 + 1.8 = 7.3 \(\approx\) = 7

Median = \(\sum f\) = 23+1 = 24Th = \(\frac{24}{2}\) = 12TH position is 8 = 3 Range = 5 - 0 = 5 (3, 5)

set = 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 multiples of 3 = 6, 9, 12, 15 multiples of 4 = 4, 8, 12 Prob(multiples of 3 or 4) = Prob(multiple of 3) + pro(multiple of 4) - pro(3 ∩ 4) = \(\frac{4}{12} + \frac{3}{12} = \frac{7}{12} - \frac{1}{2}\) = \(\frac{4 + 3 - 1}{12}\) = \(\frac{1}{2}\)

Modal class ≈ 40.5 That is, the class having the highest frequency.

∑f = 2 + 2 + 8 + 4 + 4 = 20