Answer: Option A. -> 14
:
A
P(It lands on R)= P( S to R)+ P(Q to R)
From Q it can move in 2 directions.
P(Landing on R)= $\frac{1}{6}\times 1+\frac{1}{6}\times \frac{1}{2}=\frac{1}{4}$

Answer: Option A. -> 2
:
A
P(All different)= $\left({}^x{C}_1\right)\times \left({}^{2x}{C}_1\right)\times \left({}^{3x}{C}_1\right)/\left({}^{6x}{C}_3\right)$
P(E)= $\left(6x^3\right)\times \frac{6}{6x}\times \left[\right(6x-1)\times (6x-2\left)\right]=\frac{3}{10}$
or, $\frac{3x^2}{\left[\right(6x-1)\times (6x-2\left)\right]}=\frac{3}{10}$
$10x^2=18x^2-9x+1$
or, (x-1)(8x-1)=0
x can not be a fraction hence x=1
Total number of white balls= 2x=2

Answer: Option A. -> 17105
:
A
P(A U B)= P(A)+ P(B)- P(A intersection B)
$\frac{2}{3}=\left(\frac{2}{5}\right)+\left(\frac{4}{7}\right)-x$
$x=\frac{17}{105}$

Answer: Option A. -> 2nCn22n 
:
A
We know that the number of sub-sets of a set containing n elements is $2^n$.
Therefore the number of ways of choosing A and B is $2^n.2^n=2^{2n}$
We also know that the number of sub-sets (of X) which contain exactly r elements is ${}^n{C}_r.$
Therefore the number of ways of choosing A and B, so that they have the same number elements is
${(}^n{C}_0{)}^2+{(}^n{C}_1{)}^2+{(}^n{C}_2{)}^2+....+{(}^n{C}_n{)}^2{=}^{2n}{C}_n$
Thus the required probability = $\frac{{}^{2n}{C}_n}{2^{2n}}.$

Answer: Option D. -> None of the above
:
D
P(A)= $\frac{4}{5}$
P(R)=$\frac{1}{3}$
P(E)= P(A'B)+P(AB')
P(E)=$\frac{1}{5}\times \frac{1}{3}+\frac{4}{5}\times \frac{2}{3}$
P(E)=$\frac{1}{15}+\frac{8}{15}$
P(E)= $\frac{9}{15}$
P(E)= $\frac{3}{5}$

Answer: Option A. -> (n−1)(52−n)(51−n)50×49×17×3
:
A
Here the least number of draws to obtain 2 aces are 2 and the maximum number is 50 thus n can take value from 2 to 50.
Since we have to make n draws for getting two aces, in (n – 1) draws, we get any one of the 4 aces and in the $n^{th}$draw we get one ace. Hence the required probability
$=\frac{{}^4{C}_1{\times }^{48}{C}_{n-2}}{{}^{52}{C}_{n-1}}\times \frac{3}{52-(n-1)}$
$=\frac{4\times \left(48\right)!}{(n-2)!(48-n+2)!}\times \frac{(n-1)!(52-n+1)!}{\left(52\right)!}\times \frac{3}{52-n+1}$
$=\frac{(n-1)(52-n)(51-n)}{50\times 49\times 17\times 3}$(on simplification).

Answer: Option B. -> 713
:
B
Total no. of outcomes = 52
Number of black cards (Spade+Club) in a pack of '52' cards = 26
Number of 'Kings' in a pack of cards = 4
Number of 'Black Kings' that have already been includedin the number of black cards= 2
$\therefore$ Number of favourable outcomes $=26+4-2=28$
$\therefore$ Probability(getting a black or king) $=\frac{28}{52}=\frac{7}{13}$

Answer: Option A. -> 1118
:
A
Given,
Number of green balls = 5
Number of black balls = 6
Number of white balls = 7
Total numberof 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(balldrawn isnotwhite) = $\frac{11}{18}$
$\therefore$ Probability that the ball drawn is not white is $\frac{11}{18}$ .
Alternate Method:
P (balldrawn is white) = $\frac{7}{18}$
By complementary event formula,
P( balldrawn is white) +P( balldrawn is not white) = 1
$\Rightarrow$ P( balldrawn is not white)
$=1-\text{P( balldrawn is white)}$
$=1-\frac{7}{18}=\frac{11}{18}$
$\therefore$ Probability that the balldrawn is not white is $\frac{11}{18}$.

Answer: Option B. -> 125
:
B

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 = $\frac{area\left(innermostcircle\right)}{area\left(outermostcircle\right)}$
$=\frac{\pi {.4}^2}{\pi {20}^2}$
$=\frac{1}{25}$

Answer: Option C. -> 34
:
C
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 onehead 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 onehead and tail) = $\frac{6}{8}$ =$\frac{3}{4}$
$\therefore$ The probability of getting at least onehead and a tail is$\frac{3}{4}$.