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No of women wearing both purple and white shoes = 21 + 9 = 30 women. No of women wearing neither purple nor white shoes = 51 - 30 = 21 women \(\therefore\) Probability of picking a woman wearing neither purple nor white shoes = \(\frac{21}{51}\) = \(\frac{7}{17}\).

Total number of balls = 5 + 3 + 4 = 12 balls P(first ball is red) = \(\frac{5}{12}\) P(second ball is red) = \(\frac{4}{11}\) \(\therefore\) P(both balls are red) = \(\frac{5}{12} \times \frac{4}{11}\) = \(\frac{5}{33}\)

Total number of balls = 5 + 3 + 4 = 12 balls P(one ball is green and the other is blue) = P(first ball is green and second blue) + P(first ball is blue and the second green) = \(\frac{3}{12} \times \frac{4}{11} + \frac{4}{12} \times \frac{3}{11}\) = \(\frac{2}{11}\)

The probability that an event will certainly occur + the probability that an event will certainly not occur = The probability that an event will certainly occur = 1 - the probability that an event will certainly not occur The probability that an event will certainly not occur = 1 + The probability that an event will certainly occur Option C is true.

MODE \(\approx\) 20.5 + 3.5 = 24years

The interquartile range formula is the first quartile subtracted from the third quartile: IQR = Q3 – Q1.

To calculate Mean: add up all the numbers ⇒ 3+5+8+k+14+17 = 47 + k then divide by how many numbers there are ⇒ 6 The Mean is already given as 11 11 = (47 + k) / 6 Cross multiply to solve for k : k = (6 X 11) - 47 k = 19

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

To find the probability that two beads drawn (with replacement) are of the same color, we consider two cases: both red and both white. Beads = 5, White beads = 7, Total beads = 12 Both Red: \(P(\text{Red}) = \frac{5}{12}, \quad P(\text{Both Red}) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}\) Both White: \(P(\text{White}) = \frac{7}{12}, \quad P(\text{Both White}) = \frac{7}{12} \times \frac{7}{12} = \frac{49}{144}\) Total Probability of Same Color: \(P(\text{Same Color}) = \frac{25}{144} + \frac{49}{144} = \frac{74}{144} = \frac{37}{72}\) The probability that the two beads are the same color is \( \frac{37}{72}\).

T H T T, T T, H H H , T H ,H Pro(one head) = \(\frac{3}{4}\) = \(\frac{3}{4}\)

x = no. of ppl that like none. no. of ppl that like both Traditional and Modern music, which is equal to 3x Modern Music = 60 - 3x Traditional Music = 50 - 3x 60-3x + 50 - 3x + 3x + x = 94 110 - 3x + x = 94 -2x = 94 - 110 =>-2x = -16, this x = 8. Members that like only one music: = 60 - 3x + 50 - 3x = 60 - 3[8] + 50 - 3[8] = 60 - 24 + 50 - 24 = 36 + 26 = 62 Members that like only one type of music = 62

A perpendicular bisector always divides a line segment through its midpoint. The term bisect itself means dividing equally or uniformly.

P (X or Y) = P(X) + P(Y), when they are independent as given. 0.7 = 0.4 + P(Y) P(Y) = 0.7 - 0.4 = 0.30

Mean = \(\frac{{0 + (x+2) + (3x+6) + (4x+8)}}{4}\) = 4 => 0 + (x+2) + (3x+6) + (4x+8) = 16 8x + 16 = 16 x = 0 Now prepare a table showing the deviation of each of 0, (x+2), (3x+6), and (4x+8). Adding the deviations will give 12. Thus M.D = \(\frac{12}{4}\) = 3

U = {1, 2, 3, 4, 5,..., 20} E1 = {3, 6, 9, 12, 15, 18} E2 = {4, 8, 12, 16, 20} P(E\(_2\)) = \(\frac{5}{20}\) P(not E\(_2\)) = 1 - (\(\frac{5}{20}\)) = \(\frac{15}{20} = \frac{3}{4}\)