If A is selected, then two more students need to be selected. This can be done in ${}^5$C${}_2$ ways= 10

If C is also selected, only one more out of the remaining 4 needs to be selected.
That can be done in 4 ways.
Thus, Probability =$\frac{4}{10}$ = 0.4

Probability of a success= 0.3

Probability of a failure= 0.7

Required Probability = 0.3 $\times$ 0.3 $\times$ 0.7 $\times$ ${}^3$C${}_1$ = 0.189

The probability of getting no six is ${\left(\frac{5}{6}\right)}^3$ = $\frac{125}{216}$

Hence the probability of getting at least one six is 1 - $\frac{125}{216}$ = $\frac{91}{216}$

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 $\frac{2}{5}$

 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 = $\frac{{}^4{C}_3{\times }^{11}{C}_5}{{}^{15}{C}_3}\times \frac{1}{{}^7{C}_1}$

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 = $\frac{24}{90}$= $\frac{4}{15}$

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 – $\frac{1}{12}$= $\frac{11}{12}$ .

The total possibilities are 2 $\times$ 2 $\times$ 2 = 8 (each customer has two possibilities).
These are AAA, AAB, ABA, BAA, BBA, BAB, ABB, BBB.
The favourable outcomes are only 4 (AAA, AAB, ABA, BAA).
Thus the probability = $\frac{4}{8}$ = $\frac{1}{2}$ .

Probability of getting an even number of the first dice and the second dice is= $\frac{3}{6}$ $\times$ $\frac{3}{6}$ = $\frac{1}{4}$.

 No matter what sign Arpit throws, there is one sign Rita could throw that would beat it, one that would tie, and one that would lose. Rita is equally likely to throw any one of the three signs. Therefore, the Probability that Arpit will win is $\frac{1}{3}$

Probability that Rita will win = $\frac{1}{3}$

Probability of a tie = $\frac{1}{3}$.

Probability that Arpit will win = $\frac{1}{3}$ 

The correct answer is (d).