18 marbles

\((i){1\over6}(ii){1\over3}(iii){1\over6}\)

(i)\(7\over9\)
(ii)\(7\over9\)
(iii)\({4\over9}\)

Number of red marbles = 10
Number of blue marbles = 5
Number of green marbles = 7
\(\therefore\)Total number of marbles = 10+5+7=22
Now, (i) P (red marbles) = \(\frac{10}{22}=\frac{5}{11}\)
(ii) P(green marbles) = \(\frac{7}{22}\) 
and (iii)P (not blue marbles) = 1 — P (blue marbles) 
=1-\(\frac{5}{22}=\frac{17}{22}\)​​​​​​​

When all jacks (4) queens (4) and king (4) are removed then, sample space =52-12=40
(i)P (card has a value 7)=\({4\over40}={1\over10}\)
(ii)P (card has a value >7)=\({12\over40}={3\over10}\)
(iii)P (card has a value <7) =\({24\over40}={3\over5}\)

When two dice are thrown simultaneously, then sample space contains (6 x 6 = 36) outcomes.
(i) Product of the numbers on the top is 6 [(1 ,6), (2,3), (3,2), (6,1)] i.e., 4 cases
P (product is 6) = \(\frac{4}{36}=\frac{1}{9}\)
(ii)Product of the numbers on the top is 12 [(2,6), (3,4), (4,3), (6,2)] i.e., 4 cases 
P (product is 12) = \(\frac{4}{36}=\frac{1}{9}\)
(iii)Product of the numbers on the top is 7 i.e., no case  
P (product is 7) = \(\frac{0}{36}=0\) 

When two dice are thrown simultaneously, then sample space contains (6 X 6 = 36) outcomes.
(i) Sum of numbers is 7 [(6,1), (5,2), (4,3), (3,4), (2,5), (1 ,6)]
\(\therefore\) P (sum 7) = \(\frac{6}{36}=\frac{1}{6}\) 
(ii) Sum of numbers is a prime number
[(1,1), (1,2), (1,4), (2, 1), (2,3), (3,2), (4,1),(1,6),(2,5), (3,4), (4,3), (5,2), (5,6), (6,1),(6,5)]
\(\therefore\)P (sum is a prime number) =\(\frac{15}{36}=\frac{5}{12}\) 
(iii) Sum of numbers is 1 i.e., no case
\(\therefore\)P (sum is 1) =\(\frac{0}{36}=0\) 

(a) When three coins are tossed simultaneously, then the number of possible outcomes = 8,
(i.e. , HHH, HTH, THH, TTH, HHT, HTT, THT, TTT)
Number of favourable outcomes (three heads) = l, (i.e., HHH)
\(\therefore\)Required probability = \(\frac{1}{8}\) 
(b) Number of favourable outcomes (exactly 2 3, (i.e., HTH, THH, HHT)
\(\therefore\)Required probability =\(\frac{3}{8}\) 
(c) Number of favourable outcomes (at least two heads) = 4, (i.e., HTH, THH, HHT, HHH)
Required probability =\(\frac{4}{8}=\frac{1}{2}\) 

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 outcomes = 36
(i) let A = a prime number on each die.
Favourable outcomes of event A are (2, 2), (2,3), (2, 5), (3, 3), (3, 2), (3, 5),(5,2),(5,3),(5,5)
\(\therefore\)P(A)=\(\frac{9}{36}=\frac{1}{4}\)
(ii) let B = a total of 9 or 11
Favourable outcomes of event B are 
(3,6),(4,5),(5,4),(5,6),(6,3),(6,5)
\(\therefore\)P(B)=\(\frac{6}{36}=\frac{1}{6}\)

x can take the values 1, 2, and 3
y can take the values 1,4,9.
xy can take one of these values
\([1\times 1,\quad 1\times 4,\quad 1\times 9,\quad 2\times 1,\quad 2\times 4,\quad 2\times 9,\quad 3\times 1,\quad 3\times 4,\quad 3\times 9]\)
= [ 1, 4, 9, 2, 8, 18, 3, 12, 27]
In none of these cases xy = 10
\(\therefore \) xy = 10 is an impossible event 
\(\Longrightarrow \ P(xy\ =\ 10)\ =\ 0\ \ \)

Average of 3, 5, 5, 7, 7, 7, 9, 9, 9, 9 is
= \(\frac{3+5+5+7+7+7+9+9+9+9}{10}\)
=\(\frac{3+10+21+36}{10}\) = \(\frac{70}{10}\)
P(the selected number is the average i.e. 7) = \(\frac{3}{10}\)

Total number of cards = 100
(i) Number of cards bearing even number = 50
\(\therefore \) Probability of drawing an even number = \(\frac { 5 }{ 100 } =\frac { 1 }{ 2 } \)
(ii) Total cards with number less than 14= 12
\(\therefore \) Probability of drawing a number less than 14 = \(\frac { 12 }{ 100 } =\frac { 3 }{ 25 } \)
(iii) Total perfect squares{4,9,16,25,36,49,64,81,100} = 9
\(\therefore \)  Probability of drawing a perfect square = \(\frac { 9 }{ 100 } \)  
(iv) Prime numbers less than 20 are 2,3,5,7,11,13,17,19
\(\therefore \) Probability of drawing a prime number less than 20 is \(\frac { 8 }{ 100 } =\frac { 2 }{ 25 } \)

Total number of cards = 52
(i) Number of ace = 4
\(\therefore\)  Probability of drawing an ace = \(\frac{4}{52}=\frac{1}{13}\)
(ii) There is only one '2' of spades
\(\therefore\) Probability of drawing a '2' of spade = \(\frac{1}{52}\)
(iii) '10' of a black suit there are two cards
\(\therefore\)  Probability of drawing 10 of black suit = \(\frac{2}{52}=\frac{1}{26}\)

Number of possible outcome = 36
Let A be the event for getting number whose product is a perfect square.
A = (1,1), (1,4), (2,2), (3,3)(4,1), (4,4), (5,5) (6,6)
Number of outcome of A = 8. Hence, P(A) = \(\frac{8}{36}=\frac{2}{9}\)