P(winning the game) = 0.08
P(losing the game) = 1- 0.08 = 0.92

Product of 6 are (1, 6); (2,3); (6, 1); (3,2)
No. of possible out comes = 4
Total number of chances = 6\(\times\)6 = 36
P (Product of 6) = \(\frac{4}{36} \) = \(\frac{1}{9}\).

There are 365 days in a non-leap year.
\(\because\) 365 days = 52 weeks + 1 day
\(\therefore\) One day can be M, T, W, Th, F, S, S = 7
\(\therefore\) P(53 Mondays in non-leap year) = \(\frac{1}{7}\).

Possible outcomes are 4, 9, 16, 25, 36, 49, i.e. 6.
\(\Rightarrow\) P(perfect square number) =  \(\frac{6}{48}​​\)or \(\frac{1}{8}​​\)

S = {1, 2, 3, 4, 5, 6}
n(S) = 6
A = {1, 2}
n(A) = 2
\(P\left( A \right) =\frac { n\left( A \right) }{ n\left( S \right) } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)

Here, total number of face cards are 12
Number of queens are 4
Therefore Required probability = \(\frac { 4 }{ 12 } =\frac { 1 }{ 3 } \)

Probability of sure event = 1

\(\frac{1}{3}\)

Possible outcomes are HH, HT, TH, TT
\(\Rightarrow\) Number of total outcomes = 4
Let A = Atleast one head
Favourable outcomes of event A are HH, HT or TH
\(\therefore\) Number of favourable outcomes = 3
 P(A) = \(\frac{3}{4}\)

Total number of cards = 52 and number of Jack of hearts  =1
\(\therefore\)  Probability of drawing a Jack of hearts = \(\frac{1}{52}\)

probability that it will not rain tomorrow
 = 1 - Probability that it will rain tomorrow = 1 - 0.85 = 0.15

When two coins are tossed simultaneously
Total number of number of outcomes = {HH, HT, TH, TT}
Total number of outcomes = 4
Favourable outcomes = {HT, TH} = 2
Probability of getting exactly one head = \(\frac{2}{4}=\frac{1}{2}\)

When we toss a coin,we get either head or tail which are equally likely outcomes. Hence, the result of the toss of a coin is completely unpredictable or unbiased. So, tossing a coin is a fair way of deciding.

Total number of outcomes = 30
(i) let A = getting a number greater than 3 on each die.
Favourable outcomes of event A are
(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)
\(\therefore P(A)=\frac{9}{36}=\frac{1}{4}\)
(ii) let B = getting a total 6 or 7 of the numbers on two dice.
Favourable outcomes to event B are
(1, 5), (1, 6), (2, 4), (2, 5), (3, 3), (3, 4), (4, 2), (4, 3), (5, 1), (5, 2), (6, 1)
\(\therefore P(B) = \frac{11}{36}\)

Total number of playing cards = 52
(i) Favourable cases for a king or a jack are 8 (4 kings + 4 jacks)
\(\therefore\)  Probability of drawing a king or a jack = \(\frac{8}{52}=\frac{2}{13}\)
(ii) Favourable cases for a non-ace are 48 (52 cards - 4 aces)
\(\therefore\)  Probability of drawing a non-ace = \(\frac{48}{52}=\frac{12}{13}\)
(iii) Favourable cases for a red cards are 26 (13 hearts + 13 diamonds)
\(\therefore\)  Probability of drawing a red card = \(\frac{26}{52}=\frac{1}{2}\)
(iv) Favourable cases for neither a king nor a queen are 44 (52 cards - 4 kings - 4 queen)
\(\therefore\)  Probability of drawing neither a king nor a queen = \(\frac{44}{52}=\frac{11}{13}\)

(i) Total number of cards = 5
\(\therefore\) Number of all possible outcomes = 5
P (picking a queen card)= \(\frac{1}{5}\)
[\(\because\) as there is only one queen]
(ii) Suppose a queen is drawn and put a side. then, four cards are left namely, ten, jack, king and ace of diamonds.
Now, number of all possible outcomes = 4
(a) P(the second card picked up is an ace) = \(\frac{1}{4}\)
(b) P (the second card picked up is a queen)
\(=\frac{0}{4}=0\)   [\(\because\) queen is drawn before]

When two coins are tossed simultaneously, all possible outcomes are {HH, HT, TH, TT}.
Total number of sample space is n(S)=4
(i) Let E₁=Event of getting no head={TT}
n(E₁)=1
∴ P (getting no head)\(=\frac { n(E_{ 1 }) }{ n(S) } =\frac { 1 }{ 4 } \)
(ii) Let E₂=Event of getting at most one tail
={HT, TH, HH}
n(E₂)=3
∴ P (getting atmost one tail)\(=\frac { n(E_{ 2 }) }{ n(S) } =\frac { 3 }{ 4 } \)
(iii) Let E₂=Event of getting one tail={HT, TH}
n(E₃)=2
∴ P (getting one tail)\(=\frac { n(E_{ 3 }) }{ n(S) } =\frac { 2 }{ 3 } =\frac { 1 }{ 2 }\)
(d) Let E₄=Event of getting one head and one tail
={HT, TH}
n(E₄)=2
∴ P (getting one head and one tail)\(=\frac { n(E_{ 4 }) }{ n(S) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 }\)

Total number of pieces = 8 + 10 = 18
(i) No.of triangles = 8. Hence, P(triangle is lost) = \(\frac{8}{18}=\frac{4}{9}\)
(ii) No.of squares = 10. Hence, P(square is lost) = \(\frac{10}{18}=\frac{5}{9}\)
(iii) No.of squares of blue colour = 6. So, P(square of blue colour is lost) = \(\frac{6}{18}=\frac{1}{3}\)
(iv) No.of triangles of red colour = 8 - 3 = 5. So, P(triangle of red colour is lost) = \(\frac{5}{18}\)