1. 

Here S = {HH, HT, TH, TT}
Let E = event of getting at least one tail = {HT, TH, TT}
∴ P(E) = n(E)/n(S) = 3/4
2. 

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}
∴ P(E) = n(E)/n(S) = 4/8 = 1/2.
3. 

In two throws of a die, n(S) = (6 x 6) = 36
Let E = event of getting a sum 9 = {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ P(E) = n(E)/n(S) = 4/36 = 1/9.
4. 

Let S be the sample space. Then,
n(S) = 52C2 = (52 x 51)/(2 x 1) = 1326
Let E = event of getting 2 aces out of 4.
∴ n(E) = 4C2 = (4 x 3)/(2 x 1) = 6
∴ P(E) = n(E)/n(S) = 6/1326 = 1/221.
5. 

Let S be the sample space. Then,
n(S) = ^{52}C_{2} = (52 x 51)/(2 x 1) = 1326
Let E = event of getting 1 diamond and 1 heart.
∴ n(E) = number of ways of choosing 1 diamond out of 13 and 1 heart out of 13 = (^{13}C_{1} x ^{13}C_{1}) = (13 x 13) = 169.
∴ P(E) = n(E)/n(S) = 169/1326 = 13/102.
6. 

Let S be the sample space. Then,
n(S) = number of ways of drawing 3 marbles out of 12 ^{12}C_{3}
n(S) = (12 x 11 x 10)/(3 x 2 x 1) = 220
Let E be the event of drawing 3 marbles of the same color.
Then, E = event of drawing (3 balls out of 5) or (3 balls out of 4) or (3 balls out of 3)
⇒ n(E) = (^{5}C_{3} +^{ 4}C_{3} + ^{3}C_{3}) = (^{5}C_{2} + ^{4}C_{1} + 1) = [(5 x 4)/(2 x 1)] + 4 + 1 = 15
⇒ P(E) = n(E)/n(S) = 15/220 = 3/44.
∴ Required probability = (1 – 3/44) = 41/44.
7. 

Let S be the sample space. Then,
n(S) = number of ways of drawing 3 balls out of 15 = ^{15}C_{3} = (15 x 14 x 13)/(3 x 2 x 1) = 455.
Let E = event of getting all the 3 white balls.
∴ n(E) = ^{5}C_{3} = ^{5}C_{2} = (5 x 4)/(2 x 1) = 10.
∴ P(E) = n(E)/n(S) = 10/455 = 2/91.
8. 

Total number of balls = (2 + 3 + 2) = 7
Let S be the sample space. Then,
n(S) = number of ways of drawing 2 balls out of 7 = ^{7}C_{2} = (7 x 6)/(2 x 1) = 21.
Let E = event of drawing 2 balls, none of which is red.
∴ n(E) = number of ways of drawing 2 balls out of (2 + 3) balls
= ^{5}C_{2} = (5 x 4)/(2 x 1) = 10
∴ P(E) = n(E)/n(S) = 10/21.
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