The maximum total possible = 18

There are two options for Vijay

1) A+B+C=17 Possibilities = 6,6,5

Number of combinations = $\frac{3!}{2!}$ =3

2) A+B+C=18Possibilities = 6,6,6

Number of combinations = 1

Required Probability= $\frac{4}{216}$ = $\frac{1}{52}$.

Option (c)

Probability of all three balls being blue = $\frac{{}^5{C}_3}{{}^9{C}_3}$ = $\frac{5}{52}$

Odds in favour of being blue =$\frac{5}{(42-5)}$= 5 : 37

Out of the 12 nuts, 4 are defective and out of the 24 bolts, 8 are defective the probability of picking up both non-defective tools = $\frac{24}{36}$ $\times$ $\frac{23}{135}$ = $\frac{49}{105}$

7 can be drawn in the following ways $\to$ 1 + 6, 2 + 5, 3 + 4, 6 + 1, 5 + 2 and 4 + 3.

Total number of ways = 6

Total number of possibilities = 6 $\times$ 6

Probability of not drawing a seven = $\frac{(36-6)}{36}$ = $\frac{30}{36}$

Odds against drawing a 7 is = 30 : 6 or 5 : 1.

 Option (d)

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

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)

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)

An integer can end with any of the ten’s digits (0, 1, 2 ... 9) out of which if it ends with one of the four (0, 1, 5, 6), the required condition will be satisfied. The probability of an integer ending with 0 or 1 or 5 or 6 is $\frac{4}{10}$ = $\frac{2}{5}$ Now the probability of second integer also ending with the digit that has come in the unit’s place of the first integer is $\frac{1}{10}$

Therefore, the required probability = $\left(\frac{2}{3}\right)$ $\times$ $\left(\frac{1}{10}\right)$ = $\frac{1}{25}$

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}$ =$\frac{1}{8}$.

 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)