Answer: Option C. -> 0.048
:
C
Option (c)
If the first square chosen is one of the 4 corner squares, the second square can be chosen in 1 way = 4 $\times$1. If the first square chosen is one of the squares on the sides (other than corners) = 24, the second square can be chosen in 2 ways = 24 $\times$ 2
If the first square is any of the middle squares = 36, the second square can be chosen in 4 ways = 36 $\times$ 4. Total number of ways = 4 + 48 + 144 = 196
Number of ways in which 2 random squares can be selected in a chess board = 64 $\times$ 63
Required probability = $\frac{196}{(64\times 63)}$ = 0.048

Answer: Option A. -> 4C3×11C515C3 × 17C1
:
A
Lets consider upto the 9${}^{th}$apple. The first 8 apples should have 3 rotten ones and remaining 5 good ones. This can be chosen in ${}^4$C${}_3$× ${}^{11}$C${}_5$ ways.
Total number of selections of 8 apples out of 15 apples is ${}^{15}$C${}_8$ .
The last apple is the only rotten one left, which can be selected in 1 way.
Total number of ways of selecting that 1 apple from remaining 15 – 8 = 7 apples = ${}^7$C${}_1$ ways
Total probability = $\frac{{}^4{C}_3{\times }^{11}{C}_5}{{}^{15}{C}_3}\times \frac{1}{{}^7{C}_1}$

Answer: Option B. -> 649
:
B
Total number of ways in which centres can be allocated = 7${}^4$. Two centres can be chosen in ${}^7$C${}_2$ ways. The number of ways they can accommodate 4 students = 2${}^4$
But this includes the two cases where one centre is having all the four students. So, no. of ways = 2${}^4$ – 2 = 14. So, total number of favourable ways = ${}^7$C${}_2$ $\times$ 14
So, probability = ${}^7$C${}_2$ $\times$ $\frac{14}{7^4}$ = $\frac{6}{49}$.
Hence option (b)

Answer: Option B. -> 1112
:
B
First find the probability of sum being greater than 10. Combination whose sum of 12 is (6, 6)
Combinations whose sum of 11 is (5, 6), (6, 5).
Therefore, there are totally 3 occurrences out of 36 occurrences that go against the given condition.
Probability whose sum of two numbers is greater than or equal to 11 = $\frac{3}{36}$ = $\frac{1}{12}$.
Hence probability whose sum of two numbers is lesser than 11 = 1 – $\frac{1}{12}$=$\frac{11}{12}$ .

Answer: Option D. -> 80243
:
D
An experiment succeeds twice as often as it fails. So, the probability of success is $\frac{2}{3}$and the probability of failure is $\frac{1}{3}$.
In the next 5 trials, the experiment needs to succeed in 4 out of the 5 trials. 4 out of 5 trials can be selected in ${}^5$C${}_4$ = 5 ways.
So, required probability = 5 ×${\left(\frac{2}{3}\right)}^4$×$\left(\frac{1}{3}\right)$= $\frac{80}{243}$
Hence option (d)

Answer: Option B. -> 25
:
B
Total sample space is 30
The number of multiples of 3 are 10 and the number of multiples of 13 are 2, out of which none of them are common.
Hence, answer is $\frac{2}{5}$

Answer: Option B. -> 724
:
B
The number of ways that no two girls are placed together is the number of ways in which 3 places marked with G are selected out of the SEVEN places.
___B___B___B___B____B____B____
This can be done in ${}^7$C${}_3$ ways.
Total no. of ways in which balls can be arranged = $\frac{10!}{(7!\times 3!)}$ = ${}^{10}$C${}_3$. So, required probability = $\frac{{}^7{C}_3}{{}^{10}{C}_3}$ = $\frac{7}{24}$.
Hence option(b)

Answer: Option C. -> 18
:
C
option (c)
For it to be an actual six two things are required to happen together.
1. He is speaking the truth. Let this be an event A.
2. The die shows 6. Let this be the event B.
We have: P(A) =$\frac{3}{4}$P(B) = $\frac{1}{6}$
The probability that both events happen together = P(A) $\times$ P(B) =$\frac{3}{4}$$\times$$\frac{1}{6}$=$\frac{1}{8}$.

Answer: Option C. -> 14
:
C
Probability of getting an even number of the first dice and the second dice is= $\frac{3}{6}$ $\times$ $\frac{3}{6}$ = $\frac{1}{4}$.

Answer: Option B. -> 415
:
B
Sample space is 90 numbers, of which , $\frac{90}{3}$ = 30 are multiples of 3. $\frac{90}{5}$= 18 are multiples of 5, of which every 1 out of 3 are multiples of both 3 and 5.
Probability that a two-digit number is a multiple of 3 and not a multiple of 5 = 30-6 = $\frac{24}{90}$= $\frac{4}{15}$