1. Experiment:
An operation which can produce some welldefined outcomes is called an experiment.
2. Random Experiment:
An experiment in which all possible outcomes are know and the exact output cannot be predicted in advance, is called a random experiment.
Examples:
i. Rolling an unbiased dice.
ii. Tossing a fair coin.
iii. Drawing a card from a pack of wellshuffled cards.
iv. Picking up a ball of certain colour from a bag containing balls of different colours.
Details:
i. When we throw a coin, then either a Head (H) or a Tail (T) appears.
ii. A dice is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we throw a die, the outcome is the number that appears on its upper face.
iii. A pack of cards has 52 cards.
It has 13 cards of each suit, name Spades, Clubs, Hearts and Diamonds.
Cards of spades and clubs are black cards.
Cards of hearts and diamonds are red cards.
There are 4 honours of each unit.
There are Kings, Queens and Jacks. These are all called face cards.
3. Sample Space:
When we perform an experiment, then the set S of all possible outcomes is called the sample space.
Examples:
In tossing a coin, S = {H, T}
If two coins are tossed, the S = {HH, HT, TH, TT}.
In rolling a dice, we have, S = {1, 2, 3, 4, 5, 6}.
4. Event:
Any subset of a sample space is called an event.
5. Probability of Occurrence of an Event:
Let S be the sample and let E be an event.
Then, E ⊆ S.
∴ P(E) = n(E)/n(S).
6. Results on Probability:
i. P(S) = 1
ii. 0 ≤ P(E) ≤ 1
iii. P(Φ) = 0
iv. For any events A and B we have : P(A ∪ B) = P(A) + P(B)  P(A ∩ B)
v. If Ā denotes (notA), then P(Ā) = 1  P(A).
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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