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Sure Event is a type of event that has a 100% probability of taking place. It means all elements in the sample space indicate such types of events.
- The sure event has a probability of occurrence equivalent to 1.
- The probability of an event is defined as the subset of the required sample space.
- Sample space is defined as a collection of all sets of possible outcomes.
- Probability is a branch of mathematics that deals with the occurrence of an event.
- The sum of the probability of an event taking place is equivalent to one.
- The words ' likely ', ' definite ', ' most presumably ', ' assured ' and so on include a component of a sure event.
- It will definitely rain is an example of such type of event.
- We are foreseeing precipitation dependent on our experience when it is down-poured under comparative conditions.
Key Terms: Sure Event, Probability, Events, Test Space, Occasion, Inconceivable Occasion, Sample Space, Types of Event, Outcomes
Sure Event
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Sure Event is a form of occasion that will definitely take place. It contains the entire example space. The probability of occurence of a sure event is equal to one.
- Whenever an experiment is conducted it will take place.
- It is a type of event that is opposite of impossible events.
- Impossible event is defined as an event that has no chance of occuring.
- As a result the probability of an impossible event is equal to zero.
- It is generally utilized in the fields of Physical Sciences, Commerce, Biological Sciences, Medical Sciences and so forth.
Example of Sure EventExample 1: Consider a bag which contain only red balls. When a ball is taken out from the bag randomly then taking a red ball is a sure event. Example 2: Throw a kick the bucket multiple times and note down the occasions the numbers 1,2,3,4,5,6 come up. Discover the upsides of the accompanying divisions : (Number of times 1 comes up )/(Total number of times the kick the bucket is tossed )
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Sure Event
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Test Space
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Test Space is defined as the arrangement of the multitude of potential results is called test space. It is generally addressed with the Greek letter Ω
- Test space is processed with assertion.
- The rudimentary occasions are diverse in every issue.
Example of Test SpaceExample 1: We tossed a dice ... or We tossed a coin ... First, we need to realize what results might come out:
Example 2: "We tossed a dice. What is the likelihood of getting a four?" What we need to consider, for this situation, is that "it comes a four," is the thing that we call occasions, which are subsets of the example space. Accordingly, for this situation our example space is Ω = { 1 , 2 , 3 ,4 ,5 , 6} and our prosperity is A = " get a four " = {4} |
Definite Occasion
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A definite occasion is the one that contains the entire example space. They are marked by certain occurrences or circumstances taking place. It is an that has happened.
Example of Definite OccasionExample: For instance, in our trial of tossing dice and noticing the outcome, the occasion A = " get a number more modest than 2 or higher than or equivalent to 3 " A = { 1 ,2 ,3 ,4 , 5 ,6} = Ω |
Inconceivable Occasion
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Inconceivable occasion is also known as outlandish occasion. It is a type of event that has not occurence and indicate the empty sample space. An outlandish occasion is a contrary situation when the occasion doesn't contain any component of the example space.
Example of Inconceivable OccasionExample: For instance, the occasion A =" get a 7" on tossing a dice of six counts or B=" get a white ball " from an urn that just holds back torpedoes. Generally, these occasions are addressed with A = B = Ø which is the vacant set, or at the end of the day, we are saying that there is no conceivable outcome that fulfils the occasion. |
What is Probability?
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Probability is defined as the occurrence or likelihood of an event. It is value lies in the range of 0 to 1. In this, 0 indicates an impossible event, and 1 indicates a sure event.
- The probability of an event taking place with the required sample space is one.
- Its value cannot be negative.
- Probability is the ratio of the required number of outcomes with the total number of outcomes.
- Events can be depicted with the help of a probability tree diagram.
Probability
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| Class 10 Maths Related Concepts | ||
|---|---|---|
| Relation and Function | Analysis of Variance | Mode of Grouped Data |
| Tossing a coin or rolling a die | Types of events | Remainder Theorem |
Things to Remember
- A sure event is a type of event that has a 100% chance of occurring.
- The probability of a sure event is always one.
- It will represent all elements of sample space.
- The likelihood of a definite occasion is one, and an outlandish occasion is 0
- It lies somewhere in the range of 0 and 1
- Each elementary occasion related to an arbitrary trial has an equivalent likelihood
- The amount of the probabilities of every result in a test is 1.
Sample Questions
Ques. Assume that we flip two coins and need to know whether the occasion C =" get head in one of the two coins, or get a similar outcome in is certain. (3 marks)
Ans. Arrangement: First we make the example space Ω = {hh, ht , th, tt }
- Second, we compose our occasion. What rudimentary occasions structure the occasion?
- If we figure a little we can see that it is, indeed, a protected occasion because, if we get '" hh " or "tt" with the two coins we get something similar, and in the other two potential causes, " ht " and " the" we get ahead in of the coins.
- That is regardless of the outcome we get, C will consistently occur.
Ques. Consider the recurrence circulation table which gives loads of 38 understudies of a class. (4 marks)
| Weight ( in kg) | No of students |
|---|---|
| 31-35 | 9 |
| 36-40 | 5 |
| 41-45 | 14 |
| 46-50 | 3 |
| 51-55 | 1 |
| 56-60 | 2 |
| 61-65 | 2 |
| 66-70 | 1 |
| 71-75 | 1 |
(A) Find the likelihood that the heaviness of an understudy in the class lies in the span of 46 - 50 kg.
(B) Give two occasions in this specific circumstance, one having likelihood 0 and the other having likelihood 1.
Ans. Arrangement: (A) The absolute number of understudies is 38, and the number of understudies with weight in the stretch 46 - 50 kg is 3.
- In this way, P(weight of an understudy is in the span 46 - 50 kg) = 3/38 = 0.07.
(B) For example, consider the occasion that an understudy gauges 30 kg. Since no understudy has this weight, the likelihood of an event of this occasion is 0. Additionally, the likelihood of an understudy weighing more than 30 kg is 38/38 = 1
Ques. Fifty seeds were chosen indiscriminately from every one of 5 sacks of seeds and were held under normalized conditions good for germination. Following 20 days, the number of seeds that had sprouted in every assortment was considered and recorded as follows : (3 marks)
| bag | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Number of seeds germinated | 40 | 48 | 42 | 39 | 41 |
What is the likelihood of germination of (I) over 40 seeds in a sack? (ii) 49 seeds in a sack? (iii) over 35 seeds in a sack.
Ans. Arrangement: Total number of sacks is 5.
- Number of sacks in which more than 40 seeds sprouted out of 50 seeds is 3. P(germination of in excess of 40 seeds in a pack) = 3/5 = 0.6
- Number of sacks in which 49 seeds sprouted = 0. P(germination of 49 seeds in a sack) = 0/5 = 0.
- Number of sacks in which more than 35 seeds sprouted = 5.
Along these lines, the necessary likelihood = 5/5 = 1
Ques. A common deck of cards contains 52 cards separated into four suits. The red suits are jewels and hearts and the dark suits are clubs and spades. The cards J, Q, and K are called face cards. Assume we pick one card from the deck indiscriminately.
a) What is the example space of the investigation.
b)What is the occasion that the picked card is a dark face card. (3 marks)
Ans. Arrangement :
- The results in the example space S are 52 cards in the deck.
- Let E be the occasion that a dark face card is picked. The results in E are Jack, Queen, King of spades or clubs. Emblematically E = {J, Q, K, of spades and clubs} or E = {J♣, Q♣, K♣, J♠, Q♠, K♠}
Ques. It is given that in a gathering of 3 understudies, the likelihood of 2 understudies not having the same birthday is 0.992. What is the likelihood of 2 understudies have a similar birthday. (2 marks)
Ans. Arrangement: Probability of 2 understudies not having the same birthday is P( 2 understudies same birthday )' = 0.992
- The likelihood of 2 understudies having the same birthday P( 2 understudies same birthday) is
- P(E)' = 1 - P(E)
- P(E)=1 - P(E)'
- 1-0.992
- 0.008
Ques. A container contains 3 red balls and 5 blue balls. A ball is drawn aimlessly from the pack. What is the likelihood that the ball drawn is (I) red (ii) not red. (2 marks)
Ans. Arrangement: (i) Probability of red balls, P(red) = Favorable result/Total result
- 3/8
(ii) Likelihood of not red balls, P(red)' = 1 - P(red)
- 1 - 3/8
- 5/8
Ques. Lets confirmation
- The likelihood of a definite occasion is 1
- The likelihood of an outlandish occasion is 0
- The likelihood of an occasion lies somewhere in the range of 0 and 1. (3 marks)
Ans. Arrangement: I. In a definite occasion n(E) =n(S)
- where 'S' is the certain occasion
- Since the number of components in occasion 'E' will be equivalent to the quantity of components in example space
- By the meaning of likelihood
- P(S)=n(E)/n(S)=1
- P(S)=1
- Since E has no component n(E)=0
- from the meaning of likelihood :
- P(S)=n(E)/n(S)=0/n(S)
- P(S)=0
- Let 'S' be the example space and 'E' be the occasion
- Then, at that point,
- 0
Ques. If P(E) = 0.01, what is the probability of 'not E'. (2 marks)
Ans: It is given that P(E) = 0.01
- P(E) + P (not E) = 1
- 0.01 + P (not E) = 1 ⇒ P (not E) = 1 – 0.01
- 0.99
Thus, probability of 'not E' = 0.99.
Ques. Find the probability of getting a number less than 4 in a single throw of a die. (2 marks)
Ans. Possible outcome = {1, 2, 3}
∴ P (Getting a number < 4) = 3/6 = ½
Ques. What is the probability of not getting a 3 if you roll a dice. (3 marks)
Ans. There are 6 events that can occur. We can get 1,2,3,4,5,6 when we roll a dice.
- P(getting 3) = 1/6
- Since both the events are complimentary.
- Therefore P(getting 3) + P(not getting 3) = 1
- P(not getting 3) = 1-1/6
- P(not getting 3) = 5/6
Ques. What is the probability of not getting a white ball, if the probability of getting a white ball is 1/7. (2 marks)
Ans. Given to us that,
- P(white balls)= 1/7
- Therefore, P(not white balls) = 1- P(white balls)
- 1-1/7
- 6/7
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