SNAP Probability and Combinatorics Questions

Here n(S) = 100 (As mentioned in the problem)
Favorable cases to get the square cards = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
n(E) = 10
Probability = n(E)/n(S) =10/100=1/10

Probability of 1st ball being Non-red => $$ \frac{7}{12} $$
Probability of 2nd ball being Non-red => $$ \frac{6}{11} $$
Since, both the balls being non-red => $$ \frac{6}{11} \times \frac{7}{12} $$ => $$ \frac{7}{22} $$
Option D is correct

When IMP are always together, consider them as one group.
Number of ways of arranging IMP and other letters = 7!/2! = 7!/2!
IMP can be arranged in 3! ways.
Total number = 3!*7!/2! = 3*7!

Probability of drawing a green marble from the box is $$\frac{1}{5}$$
We know that the number of yellow marbles are twice as many as green marbles. Hence the probability of yellow marbles would be $$\frac{2}{5}$$
Since there are only 3 types of marbles in the box, so the probability of red marbles would be $$ 1 - \frac{1}{5} - \frac{2}{5} = \frac{2}{5}$$
Let the total number of marbles be k, then we have
$$\frac{10}{k} = \frac{2}{5}$$
=> k = 25

Probability of Ajit = 5/8
Probability of Ravi = 3/5
Probability of Hari = ½
Probability that the target is hit by at least 2 persons = 1 - [Prob of no hit + Prob of 1 hit].
=$$1-\left(\frac{3}{8}\times\ \frac{2}{5}\times\ \frac{1}{2}+\frac{5}{8}\times\ \frac{2}{5}\times\ \frac{1}{2}+\frac{3}{8}\times\ \frac{3}{5}\times\ \frac{1}{2}+\frac{3}{8}\times\ \frac{2}{5}\times\ \frac{1}{2}\right)$$
= $$1-\frac{31}{80}=\frac{49}{80}$$