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Important Formulae

1. Experiment:

An operation which can produce some well-defined 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 well-shuffled 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 (not-A), then P(Ā) = 1 - P(A).

Practice Questions

1.

In a simultaneous throw of two coins, the probability of getting at least one tail is:
A.1/3B.1/2
C.2/3D.3/4

Report! Answer!

Answer is: D

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.

Three unbiased coins are tossed. What is the probability of getting at least 2 heads?
A.1/4B.1/2
C.1/3D.1/8

Report! Answer!

Answer is: B

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.

What is the probability of getting a sum 9 from two throws of a dice?
A.1/6B.1/8
C.1/9D.1/12

Report! Answer!

Answer is: C

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.

From a pack of 52 cards, two cards are drawn together at random. what is the probability of both the cards being aces?
A.1/15B.25/57
C.35/256D.1/221

Report! Answer!

Answer is: D

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.

Two cards are drawn together from a pack of 52 cards. The probability that one is a diamond and one is a heart, is:
A.3/20B.29/34
C.47/100D.13/102

Report! Answer!

Answer is: D

Let S be the sample space. Then,
n(S) = 52C2 = (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 = (13C1 x 13C1) = (13 x 13) = 169.
∴ P(E) = n(E)/n(S) = 169/1326 = 13/102.

6.

A bag contains 5 red, 4 blue and 3 black marbles. Three marbles are drawn at random. What is the probability that they are not of the same colour?
A.3/44B.3/55
C.52/55D.41/44

Report! Answer!

Answer is: D

Let S be the sample space. Then,
n(S) = number of ways of drawing 3 marbles out of 12 12C3
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) = (5C3 + 4C3 + 3C3) = (5C2 + 4C1 + 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.

A box contains 4 green, 5 white and 6 black balls. Three balls are drawn at random from the box. The probability that all of them are white is:
A.1/22B.3/22
C.2/91D.2/77

Report! Answer!

Answer is: C

Let S be the sample space. Then,
n(S) = number of ways of drawing 3 balls out of 15 = 15C3 = (15 x 14 x 13)/(3 x 2 x 1) = 455.
Let E = event of getting all the 3 white balls.
∴ n(E) = 5C3 = 5C2 = (5 x 4)/(2 x 1) = 10.
∴ P(E) = n(E)/n(S) = 10/455 = 2/91.

8.

A bag contains 2 yellow, 3 green and 2 red balls. Two balls are drawn at random. What is the probability that none of the balls drawn is red?
A.10/21B.11/21
C.2/7D.5/7

Report! Answer!

Answer is: A

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 = 7C2 = (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
= 5C2 = (5 x 4)/(2 x 1) = 10
∴ P(E) = n(E)/n(S) = 10/21.

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