Given, a coin is tossed 3 times.

Total possible outcomes = {HHH, HHT, HTT, HTH, THH, TTH, THT, TTT} (where H = Heads, T= Tails)
 Total no. of possible outcomes = 8
 Favourable outcomes (getting at least one head and tail) = {HHT, HTT, HTH, THH, TTH, THT}
No. of favorable outcomes = 6
Probability of an event $E$, 
$P\left(E\right)=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$

$\Rightarrow$ P (getting at least one head and tail)  = $\frac{6}{8}$ = $\frac{3}{4}$
$\therefore$ The probability of getting at least one head and a tail is $\frac{3}{4}$.

The total number of possible outcomes(counting of the cards from 1 to 17) = 17
Favourable outcomes are 3, 6, 7, 9, 12, 14 and 15.
$\therefore$ No. of  favourable outcomes = 7     
Let $E$ be the event of getting a multiple of 3 or 7.
$\therefore$ $\text{Probability P(E)}$ = $\frac{\text{Number of outcomes favorable to E}}{\text{Number of all possible outcomes of the experiment}}=\frac{7}{17}$

Given,
Number of green balls = 5
Number of black balls = 6
Number of white balls = 7
Total number of outcomes = 5 + 6 + 7 = 18
There are 18 balls out of which 11 are not white.
$\Rightarrow$ Number of favourable outcomes = 11
$\text{Probability of an event, P(E)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
$\Rightarrow$ P( ball drawn is not white) = $\frac{11}{18}$

$\therefore$ Probability that the ball drawn is not white is $\frac{11}{18}$ .

Alternate Method:
P (ball drawn is white) = $\frac{7}{18}$
By complementary event formula, 
P( ball drawn is white) + P( ball drawn is not white) = 1 
$\Rightarrow$ P( ball drawn is not white)
$=1-\text{P( ball drawn is white)}$ 
$=1-\frac{7}{18}=\frac{11}{18}$
$\therefore$ Probability that the ball drawn is not white is $\frac{11}{18}$.

Since there are 12 face cards in a deck of 52cards, the probability of drawing a face card is $\frac{12}{52}=\frac{3}{13}$
Hence, the probability of not picking a face card = $1-\frac{3}{13}=\frac{10}{13}$

There are 3 favourable outcomes out of a total of six outcomes in this case.
Multiples of 2 $\le$ 6 are  2, 4, and 6.
Hence, the probability is $\frac{1}{2}$. 

Given, P (A) = 0.15
As, P(A) and P(not A) are complementary events, P(A) + P(not A) = 1
P (not A) = 1 – P (A) = 1 – 0.15 = 0.85

Difference in radius between 2 circles = 4cm

Let, radius of small circle be x.    
$\Rightarrow$ x + 4 + 4 + 4 + 4 = 20    (from the diagram)         
$\Rightarrow$x = 4cm

Probability that the dart hits anywhere in the small circle  = $\frac{area\left(innermostcircle\right)}{area\left(outermostcircle\right)}$

$=\frac{\pi {.4}^2}{\pi {20}^2}$
$=\frac{1}{25}$

 There are a total of 10 + 8 + 12 = 30 balls, out of which 10 are brown.
The required probability is $\frac{10}{30}$ = $\frac{1}{3}$.