Sample Space: Probability, Event, Solved Examples

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Sample space is the collection of probable outcomes of an experiment. It is also called the collection or set of potential outcomes, which is generally symbolized by S. Tossing a coin or rolling a die when we cannot guess the outcome with confidence, but we can always say all the potential possibilities. These events are referred to as random phenomena or random experiments. Such random phenomena or random experiments are often addressed by probability theory. In the case of a random experiment, the sample space is denoted by curly brackets "{}." The sample space may vary depending on the number of outcomes in an experiment, and the event is a subset of the potential outcomes. 

Key Terms: Probability, Event, Random Experiments, Outcomes, Sample Space, Probability of an event, Impossible event, Experiment or trial


What is Sample Space?

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A sample space is a set of probable results of a random experiment. The sample space is denoted by the symbol "S." An experiment's events are the subset of conceivable outcomes. Depending on the experiment, a sample space may contain a variety of outcomes. If it has a finite number of outcomes, it is referred to as discrete or finite sample spaces. The sample spaces for a random experiment are enclosed in curly brackets "{} ".

An event, represented by E, is a subset of the sample space. If an experiment's outcome is included in E, then event E has occurred.

For example, 

Tossing a Coin: If the experiment involves tossing a single coin, the sample space is the set H, T, where H indicates that the coin was heads and T indicates that the coin was tails. E=H and E=T are two conceivable outcomes. The sample space for tossing two coins is HH, HT, TH, TT, with the outcome being HH if both coins are heads, HT if the first coin is heads and the second is tails, TH if the first coin is tails, and the second heads, and TT if both coins are tails.

Rolling a Dice: There are six possible results when we roll a dice. As a result, the sample space will be S = 1, 2, 3, 4, 5, 6.

When we roll two dice together, we obtain double the number of outcomes as when we roll a single dice. We obtain 36 results when we roll two dice together (6 x 6 = 36).

So, the possible outcomes will be, S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6), (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6). 


What is Probability?

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Probability is synonymous with possibility. It is a mathematical discipline that deals with the occurrence of a random event. The value ranges from zero to one. Probability has been introduced in mathematics to estimate the probability of occurrences occurring. 

Probability is defined as the degree to which something is likely to occur. This is the fundamental probability theory, which is also utilized in the probability distribution, in which you will learn about the possible results of a random experiment. To determine the probability of a positive event occurring, we must first determine the total number of alternative possibilities.

According to the probability formula, the probability of an event occurring is equal to the ratio of the number of favorable outcomes to the total number of outcomes.

Probability of event to happen P(E) = Number of favorable outcomes/Total Number of outcomes

Probability of an Event

Assume that an event E can occur in r of n probable or feasible equally likely ways. The event's probability of occurring or succeeding is then represented as:

P(E) = r/n

The probability that the event will not occur, often known as its failure, is represented as follows:

P(E’) = (n-r)/n = 1-(r/n)

E' denotes that the event will not take place.

As a result, we can now state;

P(E) + P(E’) = 1

This implies that the sum of all probability in every random test or experiment equals 1.


Difference Between Event and Sample Space

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Even though both a sample space and an event are enclosed in curly braces "{}", there is a difference between both the terms. When we roll a die, we obtain sample space as 1, 2, 3, 4, 5, 6, but an event will represent either a set of even numbers like 2, 4, 6 or a set of odd numbers like 1, 3, 5.


Things to Remember

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  • An event is a collection of experimental results for which a probability is assigned. A single outcome might be a part of many different events, and various events in an experiment are not always equally likely since they can involve quite diverse groups of outcomes.
  • Because the entire number of favorable outcomes equals the whole number of conceivable outcomes, the probability of any certain event is always equal to one. An impossible event has 0 probability since it cannot occur in any situation.
  • Getting a number larger than 6 while rolling dice is an example of an impossible event.
  • The theoretical probability is defined as the likelihood of an event occurring from a sample space of known equally desirable options.
  • The empirical probability is dependent on observation and experience.

Sample Questions

Ques. When is an experiment referred to as a random experiment? (2 Marks)

Ans. A random experiment must meet the following two requirements:

  • It has several conceivable outcomes.
  • The outcome cannot be predicted in advance.

Ques. Make a list of the sample space for a coin that has been tossed three times. (2 Marks)

Ans. A coin has two possible outcomes: head (H) or tail (T). The total number of potential outcomes when a coin is tossed three times is the cube of two, which is eight. Thus, when a coin is tossed three times, the sample space is given by:

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

Ques. What is the likelihood of the outcomes when we toss a dice? (2 Marks)

Ans. The two outcomes of this experiment are heads and tails.

The probabilities will be: 

P(heads) = ½

P(tails) = ½

Ques: If P(A) is ?. Find P (not A). (2 Marks)

Ans. Given that: P(A) = ?

To find P(not A) = 1 – P(A)

P (not A) = 1- ?

= (5-3)/5

= ?

Therefore, P(not A) = ?.

Ques. A die is said to be "balanced" or "fair" if each side has an equal chance of winning. Assign a probability to each event in the sample space for the experiment with a single fair dice toss. Determine the occurrences' probability. "An even number is rolled," says E, and "a number larger than two is rolled," says T. (2 Marks)

Ans. With outcomes labeled according to the number of dots on the top face of the die, the sample space is the set

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

Since there are six equally likely outcomes, which must add up to 1, each is assigned probability 1/6.

Since E = {2,4,6} ,

P(E) = ? +? +? = 3/6 = 1/2

Since T= {3,4,5,6} 

P(T) = 4/6 = ?

Ques. If four whole numbers taken at random are multiplied together, then the chance that the last digit in the product is 1, 3, 5, 7 is: (2 Marks)

Ans. The last digit of the four whole numbers can be

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

The chance that any of the four numbers is divisible by 2 or 5 = 6/10 = 3/5

Hence, the chance that any of the four numbers are not divisible by 2 or 5 = 1 – 3/5 = 2/5

So, the chance that all of the four numbers are divisible by 2 or 5 = (2/5)×(2/5)×(2/5)×(2/5)

= 16/625

Ques. A locality has three properties available. Three people have applied for the homes. Each applies for a single house without consulting anybody other. The chances of all three applying for the same property is: (2 Marks)

Ans. One person can select one house out of 3 = 3C1 = 3

So, three persons can select one house out of three = 3×3×3 = 27

Thus, the probability that all the three can apply for the same house = 3/27 = 1/9

Ques. Two numbers are chosen from {1, 2, 3, 4, 5, 6} one after another without replacement. Find the probability that the smaller of the two is less than 4. (3 Marks)

Ans. Total number of ways of choosing two numbers out of six = 6C2 = (6×5)/2 = 3×5 = 15

If smaller number is chosen as 3 then greater has choice are 4, 5, 6

So, total choices = 3

If smaller number is chosen as 2 then greater has choice are 3, 4, 5, 6

So, total choices = 4

If smaller number is chosen as 1 then greater has choice are 2, 3, 4, 5, 6

So, total choices = 5

Total favourable case = 3 + 4 + 5 = 12

Now, required probability = 12/15 = 4/5

Ques. There are four machines, and it is known that two of them are defective. They are examined in a random order, one by one, until both problematic machines are detected. The likelihood of only two tests being required is? (2 Marks)

Ans. First, we choose 1 machine out of the given 4.

The probability that it is fault = 2/4 = 1/2

Now, we have to pick the second fault machine.

The probability that it is fault = 1/3

So, required probability = (1/2)×(1/3) = ?

Ques. One card is drawn from a well-shuffled pack of 52 cards. What is the probability that a card will be: (4 Marks)
(i) a diamond (ii) Not an ace (iii)a black card (iv) not a diamond

Ans. 

(i) the probability that a card is a diamond

We know that there are 13 diamond cards in a deck. Therefore, the required probability is

P( getting a diamond card) = 13/52 = ¼

(ii) the probability that a card is not an ace

We know that there are 4 ace cards in a deck.

Therefore, the required probability is

P(not getting an ace card) = 1-( 4/52)

= 1- (1/13)

=(13-1)/13

= 12/13

(iii) the probability that a card is a black card

We know that there are26 black cards in a deck.

Therefore, the required probability is

P(getting a black card) = 26/52 = ½

(iv) the probability that a card is not a diamond

We know that there are 13 diamond cards in a deck.

We know that the probability of getting a diamond card is 1/4

Therefore, the required probability is

P( not getting a diamond card) = 1- (1/4)

= (4-1)/4

= 3/4

Ques: An urn contains 6 balls of which two are pink and four are green. Two balls
are drawn at random. The probability that they are of different colors is: (3 Marks)

Ans. Given that, the total number of balls = 6 balls

Let A and B be the pink and green balls respectively,

The probability that two balls are drawn are different = P(the first ball drawn is pink)(the second ball drawn is green)+ P(the first ball drawn is green)P(the second ball drawn is pink)

= (2/6)(4/5) + (4/6)(2/5)

=(8/30)+ (8/30)

= 16/30

= 8/15

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