AP ICET 2nd May 2018 Shift-2

The mean of the given numbers = 8
The absolute deviations are : 5,3,1,1,3,5 
Sum of absolute deviations = 18
Mean deviation = 18/6 = 3

Let the first odd number be 2n+1. 
Hence, the first five odd numbers would be: 2n+1, 2n+3, 2n+5, 2n+7, 2n+9
Median of these: 2n+5

The last 5 odd numbers would be: 2n+15, 2n+17, 2n+19, 2n+21, 2n+23 
The median of these numbers: 2n+19

Hence, the difference between the medians = 2n+19 - 2n - 5 = 14

Mode is simply that entity which occurs with the highest frequency in the set of observations
Here, 21 occurs the most: 5 times out of 14 total observations.
Hence, it is the mode here. 

Let the mean of the first series be M
Hence, M = $$\frac{x_1+x_2+......+x_n}{n}$$

Now, for the second series, let its mean be M'
Hence, M' = $$\frac{3\cdot\left(3n+x_1+x_2+......+x_n\right)}{n}=9+3M$$

Now, respective deviation of the individual terms from the mean M' are:
3M - 3$$x_1$$ = 3(M-$$x_1$$)
3M - 3$$x_2$$ = 3(M-$$x_2$$)
.
.
.
3M - 3$$x_n$$ = 3(M-$$x_n$$)

Variation = $$\frac{\left[9\cdot\left(M-x_1^{ }\right)\left(M-x_1\right)+9\cdot\left(M-x_2\right)\left(M-x_2\right)+....+9\left(M-x_n\right)\left(M-x_n\right)\right]}{n}$$
Standard deviation = Square root of this Variance = 3*$$\sigma\ $$

The first 5 odd natural numbers are: 1 3 5 7 9 
Average of these = 5
Respective deviation from mean: -4 -2 0 2 4 

Variance = $$\frac{\left(Sum\ of\ squares\ of\ individual\ deviation\right)}{Number\ of\ terms\ }$$ = (16+4+0+4+16)/5 = 8

Consider the first case of Odd-Even-Odd: 
$$\frac{5}{9}\cdot\frac{4}{8}\cdot\frac{4}{7}=\frac{20}{126}$$

Consider the second case of Even-Odd-Even:
$$\frac{4}{9}\cdot\frac{5}{8}\cdot\frac{3}{7}=\frac{15}{126}$$

Hence, The net probability = $$\frac{20}{126}+\frac{15}{126}=\frac{35}{126}=\frac{5}{18}$$ 

There are a total of 14 numbers from 1 to 100 which are multiples of 7.

$$\therefore\ $$The probability that the chosen number is divisible by 7 = $$\frac{14}{100}$$ = $$\frac{7}{50}$$

Hence, the correct answer is Option C

The order of the appearance of heads and tails does not matter here.
So we can have only 4 distinct possible combinations: HHH, HTH, HTT, TTT

From this, our desired situations are HHH and HTH. 
Hence, the probability = 2/4 = 1/2