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From the table, the mode = 1. The median = 2.

\(\begin{array}{c|c} & & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline a & 1 & 1, 1 & 1, 2 & 1, 3 & 1, 4 & 1, 5 & 1, 6\\\hline & 2 & 1,1 & 2, 2 & 2, 3 & 2, 4 & 2, 5 & 2, 6 \\\hline B & 3 & 3, 1 & 3, 2 & 3, 3 & 3, 4 & 3, 5 & 3, 6\\\hline & 4 & 4 , 1 & 4, 2 & 4, 3 & 4, 4 & 4, 5 & 4, 6\\\hline Die & 5 & 5, 1 & 5, 2 & 5, 3 & 5, 4 & 5, 5 & 5, 6 \\\hline & 6 & 6, 1 & 6, 2 & 6, 3 & 6, 4 & 6, 5 & 6, 6\end{array}\) Probability of obtaining total score of 8 = \(\frac{5}{36}\)

Prob of P = \(\frac{3}{4}\) Prob of Q = \(\frac{1}{6}\) Prob of both P or Q = \(\frac{3}{4} + \frac{1}{6} = \frac{11}{12}\)

(a) P(obtaining a 3 from a die) = \(\frac{1}{6}\) P(obtaining a tail in 2 coins tossed at once) = \(\frac{1}{2}\). \(\therefore\) P(obtaining 3 and a tail = \(\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}\) (b) No of black marbles = n(B) = 7 No of red marbles = n(R) = 3 Total marble = 10 (i) P(red then black) = P(1st red, then black) or P(1st black, then red) = \(\frac{3}{10} \times \frac{7}{9} + \frac{7}{10} \times \frac{3}{9}\) = \(\frac{7}{15}\) (ii) Prob (two blacks) = \(\frac{7}{10} \times \frac{6}{9} = \frac{7}{15}\)

x = 5, \(\frac{\sum {fx}}{\sum {f}}\) = 5= \(\frac{\sum {fx}}{30}\) \(\sum {fx}\) = 5 x 30 = 150; 150 - 34 = 116 \(\sum {f}\) = 29 \(\sum {fx}\) = 116 = \(\frac{116}{29}\) = 4

\(\bar{x}\) = \(\frac{2 + 14 + 15 + 16 + 45 + 42 + 42}{40}\) = \(\frac{176}{40}\) = 4.4

No expanded working has been attached for this question yet. The verified answer is Option D.

Median = 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5,| 5, 5, |5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, Median = \(\frac{5 + 5}{2}\) = 5

Total number of marbles = 5; 1st pick = 2/5 2nd pick = 3/4; 3rd pick = 3/5; 4th pick = 2/4 ∴ Probability of picking two marbles of different colors = (2/5 x 3/4) + (3/5 + 2/4) = 12/20 = 3/5

(1/2)(1/4) = 1/8

No of defective(x) No of boxes (f) \(fx\) 4 2 8 5 7 35 6 17 102 7 10 70 8 8 64 9 6 54 \(\sum\) 50 333 \(Mean = \frac{\sum fx}{\sum f}\) = \(\frac{333}{50} = 6.66 \approxeq 6.7\) The median is the average of the 25th and 26th position = 6.

Coca-cola = 10 bottles, Fanta = 8 bottles Sprite = 6 bottles, Total = 24 P(cola-cola) = \(\frac{10}{24}\) P(not coca-cola) = 1 - \(\frac{10}{24}\) \(\frac{24 - 10}{24}\) = \(\frac{14}{24}\) = \(\frac{7}{12}\)

Total number of balls = 9, No of blue = 2, No of black = 3 \(\therefore\) No of red = 4. (a)(i) P(not blue) = \( 1 - P(blue)\) = \(1 - \frac{2}{9} = \frac{7}{9}\) (ii) P(not red) = \(1 - P(red)\) = \(1 - \frac{4}{9} = \frac{5}{9}\) (b) Picking two balls; (i) P(black) with no replacement = \(\frac{3}{9} \times \frac{2}{8} = \frac{1}{12}\) (ii) P(blue) with replacement = \(\frac{2}{9} \times \frac{2}{9} = \frac{4}{81}\)

Mean of 10 numbers = 16 The total sum of numbers = 16 x 10 = 160 Mean of 11 numbers = 18 Total sum of numbers = 11 x 18 = 198 The 11th no. = 198 - 160 = 38

Total pairs: C(4,2) = 6. Favorable (even sum): both odd (1 way: {1,3}) or both even (1 way: {2,4}) → 2 ways. P = \(\frac{2}{6}\) = \(\frac{1}{3}\).