Random variable

A random variable is a function that assigns a real number to each value in the sample space. This is produced as a result of a random experiment like tossing a coin or throwing a dice.

Example 1

When we flip a coin twice, there are four possible outcomes: both heads, heads and tails, tails and heads, or both tails. The value of x below represents the number of heads from each outcome of these four outcomes.:

Sample Space x
HH 2
HT 1
TH 1
TT 0

Example 2

Random variable Y represents the number of times we roll “5” when rolling a six-sided dice six times. Therefore, the possible values of this random variable Y, can be anything from 0 (we never roll “5”) to6 (we roll “5” every time).

Joint probability

This refers to the Probability of two or more simultaneous events and is used at many places in the Data Science field, especially when dealing with textual data where we have to find the probability of words which can occur together. The joint probability of two events is denoted as below.

P(A and B) = P (A ∩\cap B)

Where P(A) and P(B) are probabilities of events A and B respectively.

Probability for independent events

If one event is not dependent on the other, the events are said to be independent and their joint probability is calculated as below.

P(A and B) = P(A, B) = P (A ∩\cap B) = P(A) * P(B)

Where P(A) and P(B) are probabilities of events A and B respectively.

Example

A jar contains three red, five green, two blue, and six yellow marbles. A marble is chosen from the jar at random. After replacing it, a second marble is chosen. What is the probability of choosing a green and then a yellow marble?

Solution

From the above question, we can deduce the below things from the above question.

  • Number of Red Marbles = 3
  • Number of Green Marbles = 5
  • Number of Blue Marbles = 2
  • Number of Yellow Marbles = 6
  • Total Marbles = 16
  • Let A be the event that the Green marble is drawn
  • Let B be the event that the Yellow marble is drawn
  • Let C be the event that the Blue marble is drawn
  • Let D be the event that the Red marble is drawn
  • As the event of drawing marble is independent from each other we can proceed as follows.

P(A) = 516\frac{5}{16}

P(B) = 616\frac{6}{16}

P(A and B) = P(A ∩\cap B) = 516\frac{5}{16} x 616\frac{6}{16} = 15128\frac{15}{128}

Probability for Mutually exclusive events

Mutually Exclusive events are those events which cannot occur at the same time. For example when we toss a coin, the output is either Head or Tail.

  • If the two events are mutually exclusive, then the probability of both occurring is

P(A and B) = P (A ∩\cap B) = 0

  • If the two events are mutually exclusive, then the probability of either occurring is

P(A or B) = P (A ∪\cup B) = P(A) + P(B) - P (A ∩\cap B) = P(A) + P(B) - 0 = P(A) + P(B)

Where P(A) and P(B) are probabilities of events A and B respectively.

A standard deck of 52 cards plus 2 joker cards consist of the below cards.

  • There are four suits Spades, Diamonds, Hearts and Clubs.
  • There are thirteen suits of each card.
  • In each suit there are nine numbers from 2 to 10, a Jack, a Queen, a King and an Ace.
  • There are a total of 9*4 = 36 number cards, four Jacks, four Queens, four Kings and four Ace cards.

Example

  • There are four Queens in 52 deck cards, so the probability of a card being Queen is

P(Queen) = 452\frac{4}{52} = 113\frac{1}{13}

  • There are four Jacks in 52 deck cards, so the probability of a card being Jack is

P(Jack) = 452\frac{4}{52} = 113\frac{1}{13}

  • Combining the above two probabilities meaning the probability of drawing a Queen or Jack is

P(Queen) + P(Jack) = 113\frac{1}{13} + 113\frac{1}{13} = 213\frac{2}{13}

Probability for Non Mutually Exclusive Events

If the events are not mutually exclusive, their joint probability is calculated as:

P(A or B) = P(A ∪\cup B) = P(A) + P(B) - P (A ∩\cap B)

Where P(A) and P(B) are probabilities of events, A and B respectively.

Example

John is going to graduate from an industrial engineering department in a university by the end of the semester. After being interviewed at two companies he likes, he assesses that his probability of getting an offer from company A is 0.8, and his probability of getting an offer from company B is 0.6. If he believes that the probability that he will get offers from both companies is 0.5, what is the probability that he will get at least one offer from these two companies?

Solution

From the above question, we can deduce the following:

  • The probability for getting an offer from company A is 0.8, or P(A) = 0.8.

  • The probability for getting an offer from company B is 0.6, or P(B) = 0.6.

  • The probability of getting offers from both companies A and B is 0.5, or P(A ∩\cap B) = 0.5.

  • The probability for at least one offer from these companies, or P(A ∪\cup B) = ?.

Using the above formula we have

P(A or B) = P(A ∪\cup B) = P(A) + P(B) - P (A ∩\cap B) = 0.8 + 0.6 - 0.5 = 0.9

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