Sums on Probability

Sums on Probability

Introduction

Probability is a branch of mathematics that deals with the likelihood of events occurring. It is an important concept in various fields, including statistics, physics, engineering, and finance. Probability theory is used to understand and quantify uncertain outcomes and can be applied in various real-life scenarios. In this Nithra Jobs article we can discuss about probability questions and answers.

Formulas:

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:

1. Rolling an unbiased dice.

2. Tossing a fair coin.

3. Drawing a card from a pack of well-shuffled cards.

4. 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:

1. In tossing a coin, S = {H, T}

2. If two coins are tossed, the S = {HH, HT, TH, TT}.

3. 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.

Therefore, P(E) = n(E) / n(S).

6. Results on Probability:

i. P(S) = 1

ii. 0 < P (E) < 1

iii. P(o) = 0

iv. For any events A and B we have: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

v. If A denotes (not - A), then P(A) = 1 - P(A).

Solved Examples

1. An unbiased die is tossed. Find the probability of getting a multiple of 2.

Answer:

Explanation:

Here S = { 1, 2, 3, 4, 5, 6 }

Let E be the event of getting the multiple of 2 Then, E = { 2, 4, 6 }

P (E) = n (E) / n (S)

P (E) = ( 3 / 6 )

P (E) = ( 1 / 2 )

The probability is ( 1 / 2 )

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2. Three unbiased coins are tossed. What is the probability of getting at least 1 heads?

Answer:

Explanation:

Here S= { TTT, TTH, THT, HTT, THH, HTH, HHT, HHH }.

Let E = Event of getting at least one head.

E = { TTH, THT, HTT, THH, HTH, HHT, HHH }

P (E) = n (E) / n (S)

P (E) = ( 7 / 8 )

Probability is (7 / 8 )

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3. In a simultaneous throw of two dice, what is the probability of getting a doublet?

Answer:

Explanation:

In a simultaneous throw of two dice, n (S) = 6 * 6 = 36

Let E = Event of getting a doublet

n (E) = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }

P (E) = n (E) / n (S)

P (E) = ( 6 / 36 )

P (E) = ( 1 / 6 )

Probability is (1 / 6 )

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4. In a single throw of two dice, find the probability that neither a doublet nor a total of 8 will appear.

Answer:

Explanation:

n (S) = 36

A = { (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) }

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

n (A) = 6,

n (B) = 5,

n (A ∩ B) = 1

Required probability = P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

= ( 6 / 36 ) + ( 5 / 36 ) - ( 1 / 36 )

= ( 6 + 5 - 1 ) / 36

= ( 10 / 36 )

= ( 5 / 18 )

Probability is ( 5 / 18 )

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5. Two dice are tossed. What is the probability of getting sum of two numbers which is divisible by 7?

Answer:

Explanation:

Clearly, n ( S ) = ( 6 * 6 )

= 36

Let E = Event that the sum of two numbers which is divisible by 7

Then E = { ( 1, 6 ), ( 2, 5 ), ( 3, 4 ), ( 4, 3 ), ( 5, 2 ), ( 6, 1 ) }

n ( E ) = 6

P ( E ) = n ( E ) / n ( S )

= 6 / 36

= 1 / 6

The probability of getting sum of two numbers which is divisible by 7 is 1 / 6

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Conclusion

With consistent practice and an understanding of the basic concepts of probability aptitude questions, anyone can become proficient in solving probability problems. Therefore, probability is a critical concept to understand for anyone interested in mathematics, statistics, or any field that deals with uncertainty. Solve them easily with the help of probability formulas that is given above.

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