Answer: Option B. -> $$\frac{{33}}{{91}}$$
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
When four marbles are picked at random, then the probability that none is blue is
$$\eqalign{
& = \frac{{{}^{12}{C_4}}}{{{}^{15}{C_4}}} \cr
& = \frac{{12 \times 11 \times 10 \times 9}}{{15 \times 14 \times 13 \times 12}} \cr
& = \frac{{11880}}{{32760}} \cr
& = \frac{{33}}{{91}} \cr} $$

Answer: Option B. -> $$\frac{{17}}{{25}}$$
The number of exhaustive events = $${{}^{50}{C_1}}$$ = 50
We have 15 primes from 1 to 50
Number of favourable cases are 34
Required probability = $$\frac{{34}}{{50}}$$ = $$\frac{{17}}{{25}}$$

Answer: Option D. -> $$\frac{{5}}{{6}}$$
P(at least one graduate) = 1 - P(no graduates)
$$\eqalign{
& = 1 - \frac{{{}^6{C_3}}}{{{}^{10}{C_3}}} \cr
& = 1 - \frac{{6 \times 5 \times 4}}{{10 \times 9 \times 8}} \cr
& = \frac{5}{6} \cr} $$

Answer: Option B. -> $$\frac{{5}}{{9}}$$
Two balls can be picked from nine balls in $${{}^9{C_2}}$$ ways.
We select one white ball and one red ball from five white balls and four red balls.
This can be done $${{}^5{C_1}}$$ . $${{}^4{C_1}}$$ ways.
∴ The required probability
$$\eqalign{
& = \frac{{5 \times 4}}{{{}^9{C_2}}} \cr
& = \frac{{20}}{{36}} \cr
& = \frac{5}{9} \cr} $$

Answer: Option D. -> $$\frac{{4}}{{13}}$$
$$P\left( {S \cup K} \right) = $$   $$P\left( S \right) + $$  $$P\left( K \right) - $$  $$P\left( {S \cap K} \right),$$   (where S denotes spade and K denotes king.)
$$\eqalign{
& P\left( {S \cup K} \right) = \frac{{13}}{{52}} + \frac{4}{{52}} - \frac{1}{{52}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{4}{{13}} \cr} $$

Answer: Option D. -> $$\frac{{1}}{{20}}$$
The numbers can be written in 5! different ways then n(S) = 5! = 120
Let E be the event that getting 4 digit number with 1 in the unit place and 2 in the tens place when the numbers 1, 2, 3, 4 and 5 are arranged at random.
Since each desired number is ending with 1, there is only 1 way
The Second place(tens) can now be filled with 2
Here also only 1 way to fill
The next place(hundreds) can now be filled by any of the remaining 3 numbers
So, there are 3 ways to filling that place
Then, the first place can now be filled by any of the remaining 2 numbers
So, there are 2 ways to fill
Therefore n(E) = 1 x 1 x 3 x 2 = 6
Now,
$$\eqalign{
& {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{{120}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{20}} \cr} $$
Hence the required probability is $$\frac{1}{{20}}$$

Answer: Option B. -> $$\frac{{2}}{{3}}$$
Total cases = 10 + 20 = 30
Favourable cases = 20
So probability = $$\frac{{20}}{{30}}$$ = $$\frac{{2}}{{3}}$$

Answer: Option B. -> $$\frac{{3}}{{286}}$$
Total number of erasers in the box = 3 + 4 + 7 = 14
Let S be the sample space
Then, n(S) = number of ways of taking 5 out of 14
Therefore,
$$\eqalign{
& {\text{n}}\left( {\text{S}} \right) = {}^{14}{C_5} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{14 \times 13 \times 12 \times 11 \times 10}}{{2 \times 3 \times 4 \times 5}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \times 13 \times 11 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2002 \cr} $$
Let E be the event of getting all the 5 blue erasers
Therefore,
$$\eqalign{
& {\text{n}}\left( {\text{E}} \right) = {}^7{C_5} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{7 \times 6 \times 5 \times 4 \times 3}}{{2 \times 3 \times 4 \times 5}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 21 \cr} $$
Now, the required probability
$$\eqalign{
& \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{21}}{{2002}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{286}} \cr} $$

Answer: Option C. -> $$\frac{{8}}{{21}}$$
Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green = event that the ball drawn is red.
Therefore, n(E) = 8
P(E) = $$\frac{{8}}{{21}}$$

Answer: Option A. -> $$\frac{{1}}{{6}}$$
In a simultaneous throw of two dice, we have n(s) = 6 × 6 = 36
Let E = event of getting two numbers are same.
Then E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
therefore, n(E) = 6
And p(E) = p(getting two numbers are same)
$$\eqalign{
& {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{{36}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{6} \cr} $$
Hence the answer is $$\frac{{1}}{{6}}$$