Random Experiment - Probability

In probability, a random experiment is any process or activity that:

1. Can be repeated under the same conditions, and
2. Has outcomes that cannot be predicted with certainty in advance.

Even though we don’t know exactly what will happen, we do know all the possible outcomes (sample space).

Example: A coin is tossed two times and the outcomes are recorded. The possible outcomes are:

HH, HT, TH, TT

The figure below represents the outcomes in form of a tree. 

Notice that there are four possible outcomes in this experiment and none of them can be predicted beforehand.

Some Important terms related to random experiments that are used frequently in probability theory. These terms also help us describe whether an experiment is random or not.

1. Outcome: A possible result of a random experiment is called an outcome.

Example: In an experiment of drawing a card from a deck, outcomes may be Ace, 2, 3, …, King (from any suit).

2. Sample Space: The set of all possible outcomes of a random experiment is called the sample space associated with that experiment and is denoted by the symbol S.

Example: In an experiment of tossing two coins, the sample space is S = {HT, HH, TH, TT} 

3. Sample Point: Each element of the sample space is called a sample point. Or, each outcome of a random experiment is also called a sample point.

Example: For two coin tosses, each outcome (HH, HT, TH, TT) represents a sample point.

4. Trial: A trial is a single performance or execution of a random experiment.

Example: Spinning a spinner once in a game is one trial.

5. Events: An event is a subset of the sample space. It represents specific outcomes of interest.

Example: In tossing two coins, S = {HH, HT, TH, TT}

The event of “getting at least one head” is E = {HH, HT, TH}

Probability

Once we identify the sample space of a random experiment and the event of interest, we can assign probabilities to that event. For an experiment with sample space S, the goal is to assign probabilities to certain outcomes.

Example: Consider the experiment of rolling two dice and observing the ordered pair of numbers. We are interested in the event where the sum of the numbers on the two dice is equal to 6.

The first step is to consider the sample space for this experiment. 

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); }

These are all the possible outcomes of this experiment. 

|S| = 36.

The event of interest contains all outcomes where the sum is 6:

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

Thus,

∣E∣ = 5

Probability is defined as the ratio of the favorable number of outcomes and total number of possible outcomes

P(Event) = \frac{\text{Favourable Number of Outcomes}}{\text{Total Number of Outcomes}}

For this case, 

⇒ P(E) = \frac{|E|}{|S|}

⇒ P(E) = \frac{6}{36}

⇒ P(E) = \frac{1}{6}

Related Articles:

• Probability in Maths
• Events in Probability
• Dependent and Independent Event
• Conditional Probability and Independence

Sample Problems on Random Experiment

Problem 1: Write the sample space for three tosses of a coin. 

Solution: 

A coin toss gives either Head (H) or Tails(T). 

Outcome of three coin tosses will be triplets of Heads and Tails. 

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

Problem 2: Out of the following activities given below. Choose random activities: 

1. Drawing a ball from an urn containing black and white balls.
2. Multiplying 4 and 8 on the calculator. 
3. Birth of a person

Solution: 

Each random experiment/activity must fulfill the following two conditions: 

1. The experiment can be repeated a number of times under the same conditions.
2. The outcome of the experiment cannot be predicted beforehand with certainty.

Drawing a ball from an urn containing black and white balls.

This can be repeated as many times as required and also it cannot be predicted which ball will be drawn. Thus is a random activity. 

Multiplying 4 and 8 on the calculator. 

This activity can be repeated a number of times, but the result of this activity can be predicted beforehand. Thus, it is not a random activity. 

Birth of a person

Birth of a person cannot be repeated. So, it is not a random activity. 

Problem 3: Consider an experiment of rolling a die and then tossing a coin. Draw the sample space. 

Solution: 

A die when rolled given {1, 2, 3, 4, 5, 6} as result while a coin toss gives {H, T}. 

Sample space S = {( H,1); ( H,2); ( H,3); (H,4); (H,6); (H,7); ( T,1); ( T,2); ( T,3); ( T,4); ( T,5); ( T,6)}

Problem 4: Find the probability of getting 2 heads in three tosses of a coin. 

Solution: 

A coin toss gives either Head (H) or Tails(T). 

Outcome of three coin tosses will be triplets of Heads and Tails. So, the sample space is,

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

Total Number of Outcomes = 8

Counting the number of favorable outcomes, there are 3 outcomes with two heads. 

Probability = \frac{\text{Favourable Number of Outcomes}}{\text{Total Number of Outcomes}}

P = \frac{3}{8}

Problem 5: Find the probability of drawing out an ace from a well-shuffled deck of cards. 

Solution: 

A deck of cards contains 52 cards and out of which 4 are aces. The formula for probability, 

Probability = \frac{\text{Favourable Number of Outcomes}}{\text{Total Number of Outcomes}}

Favorable Number of Outcomes = 4 

Total Number of Outcomes = 52 

Probability = \frac{\text{Favourable Number of Outcomes}}{\text{Total Number of Outcomes}}

⇒ P = \frac{4}{52}

⇒ P = \frac{1}{13}

Problem 6: Find the probability of getting at least 1 head in three tosses of a coin. 

Solution: 

A coin toss gives either Head (H) or Tails(T). 

Outcome of three coin tosses will be triplets of Heads and Tails. So, the sample space is,

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

Counting the number of favorable outcomes, there are seven outcomes with atleast one head. 

Probability = \frac{\text{Favourable Number of Outcomes}}{\text{Total Number of Outcomes}}

P = \frac{7}{8}

Problem 7: A fair coin is tossed three times. What is the probability of getting exactly two heads?

Solution:

Step 1: Identify the sample space.

The sample space consists of 8 possible outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Step 2: Count favorable outcomes.

There are 3 outcomes with exactly two heads: HHT, HTH, THH

Step 3: Calculate the probability.

Probability = Number of favorable outcomes / Total number of outcomes

= 3 / 8

= 0.375 or 37.5%

Problem 8: A standard deck of 52 cards has 4 aces. If you draw two cards without replacement, what is the probability of drawing two aces?

Solution:

Step 1: Calculate the probability of drawing the first ace.

P(first ace) = 4/52 = 1/13

Step 2: Calculate the probability of drawing the second ace, given that the first card was an ace.

P(second ace | first ace) = 3/51

Step 3: Use the multiplication rule of probability for independent events.

P(two aces) = P(first ace) × P(second ace | first ace)

= (1/13) × (3/51)

= 1/221

≈ 0.0045 or 0.45%

Problem 9: Two fair six-sided dice are rolled. What is the probability that the sum of the numbers is 7?

Solution:

Step 1: List all possible combinations that sum to 7.

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

Step 2: Count the number of favorable outcomes.

There are 6 combinations that sum to 7.

Step 3: Calculate the total number of possible outcomes.

Total outcomes = 6 × 6 = 36 (each die has 6 possible outcomes)

Step 4: Calculate the probability.

Probability = Number of favorable outcomes / Total number of outcomes

= 6 / 36

= 1 / 6

≈ 0.1667 or 16.67%

Problem 10: A bag contains 3 red balls, 4 blue balls and 5 green balls. If two balls are drawn without replacement, what is the probability that both balls are the same color?

Solution:

Step 1: Calculate the probability for each color separately.

For red: P(RR) = (3/12) × (2/11) = 1/22

For blue: P(BB) = (4/12) × (3/11) = 1/11

For green: P(GG) = (5/12) × (4/11) = 5/33

Step 2: Sum the probabilities for each color.

P(same color) = P(RR) + P(BB) + P(GG)

= 1/22 + 1/11 + 5/33

= 3/66 + 6/66 + 10/66

= 19/66

≈ 0.2879 or 28.79%

Practice Problems on Random Experiment

Problem 1: In a standard deck of 52 cards, what is the probability of drawing a heart?

Problem 2: A die is rolled. What is the probability of getting an even number?

Problem 3: A bag contains 4 red, 5 blue and 3 green balls. What is the probability of drawing a blue ball?

Problem 4: Two coins are flipped. What is the probability of getting at least one head?

Problem 5: A box contains 3 defective and 7 non-defective items. If one item is selected at random, what is the probability that it is defective?

Problem 6: A single card is drawn from a standard deck of 52 cards. What is the probability of drawing a card that is either a king or a queen?

Problem 7: What is the probability of rolling a sum of 7 with two dice?

Problem 8: In a class of 30 students, 18 are female and 12 are male. What is the probability of randomly selecting a female student?

Problem 9: A bag contains 5 red, 7 yellow and 8 black marbles. What is the probability of drawing a marble that is not red?

Problem 10: A student is selected at random from a group of 15 students, where 5 are freshmen, 7 are sophomores and 3 are juniors. What is the probability that the student is a sophomore?

Problem 11: In a lottery, 4 numbers are drawn from a pool of 50 numbers. What is the probability of drawing the exact 4 winning numbers?

Problem 12: If you roll two dice, what is the probability of getting a sum greater than 9?

Problem 13: A box contains 8 white and 4 black balls. If 2 balls are drawn at random, what is the probability that both are white?

Problem 14: A restaurant offers 5 appetizers, 8 main courses and 3 desserts. What is the probability of randomly choosing an appetizer and a dessert?

Problem 15: In a deck of 52 cards, what is the probability of drawing a card that is a spade or a 10?