Given, x = {1, 2, 3, 4}
\(\Rightarrow\) n(x) = 4
y = {1, 4, 9, 16}
n(y) = 4
Totalnumber of possible products = 4\(\times\)4 = 16.
Products x.y which are less than 16 are {1\(\times\)1, 1\(\times\) 4, 1\(\times\)9, 2\(\times\)1, 2\(\times\)4, 3\(\times\)1, 3\(\times\)4, 4\(\times\)1}
n (x.y) = 8
\(\therefore\) Required probability =\(\frac{8}{16}\) =\(\frac{1}{2}\)

The possible outcomes of the experiment of tossing two coins are HH, HT, TH and TT.
(i) Outcomes favourable to Anu are HT and TH.
P (Anu getting first chance) = \(\frac{2}{4}=\frac{1}{2}\)
Outcomes favourable to Priya is HH.
\(\therefore \)  P ( Priya getting first chance) = \(\frac{1}{4}\)
Outcomes favourable to Jyoti is TT.
\(\therefore \)  P (Jyoti getting first chance) = \(\frac{1}{4}\)
(ii) No, the number of cases favourable to each one of them are not equal.
(iii) Dishonesty, as she kept two cases favourable for herself and one each for the other two friends.

(i) Number of days in a leap year=366 days
\(=(52\times 7+2)\) days=52 weeks and 2 days
Thus, a leap year always has 52 Fridays.
The remaining 2 days can be
(a) Sunday and Monday
(b) Monday and Tuesday
(c) Tuesday and Wednesday
(d) Wednesday and Thursday
(e) Thursday and Friday
(f) Friday and Saturday
(g) Saturday and Sunday.
Out of these 7 cases, we have Friday in two cases.
So, total number of possible outcomes, n(S)=7 and total number of favourable outcomes, n(E)=2
∴ P(53 Fridays)\(=\frac { n(E) }{ n(S) } =\frac { 2 }{ 7 } \)
(ii) Sample space of a die, S={1,2,3,4,5,6}
∴ n(S)=6
Let E=Event of getting an odd prime number
={3,5}
∴ n(E)=2
Hence, required probability\(=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)

\({2\over3},{8\over23}\)

\({2\over11},{2\over11},{2\over11},{1\over11},{1\over11},{1\over11},{1\over11},{1\over11}\)

(i)There are 500 possible ways to draw a card.Perfect squares of even numbers are 4, 16, 36, 64, 64, 100, 144,196, 256, 324, 400, 484.
Number of ways to draw a no. of which is a perfect square of even number =11
Probability of first player winning a prize = \(11\over 500\)
(ii)For second players No. of cards left = 500-1=499
[∵ One card has been drawn by 1st players and it has not been replaced]
Also No. of cards bearing perfect square of even number = 11
[∵ 1st players has not won ∴ Card drawn by him does not bear a perfect square even numbert]
∴ Probability of IInd player winning a prize = \(11\over 499\)

Total cards=20
(i)Probability of drawing a card bearing an odd number = \({10\over 20}={1\over 2}\)
(ii)Probability of drawing a card bearing an even number = \({10\over 20}={1\over 2}\)
(iii)Probability of drawing a prime number {2, 3, 5, 7, 11, 13, 17, 19}= \({8\over 20}={2\over 5}\)
(iv)Probability of drawing a number divisible by 3, {3, 6, 9, 12, 15, 18} = \({6\over 20}={3\over 10}\)
(v)Probability of drawing a number divisible 2 and 3 both (6, 12, 18) = \(3\over 20\)

Total outcomes
(H, H), (H, T), (T, 1), (T, 3), (T, 4), (T, 5), (T, 6)
A=(H, H), (H, T)
B=(T, 2), (T, 4), (T, 6)
C=(H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)
D=(T, 1), (T, 3), (T, 5)