Probability Mass Function

Probability Mass Function of a Discrete Random Variable

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In this section, we will understand Probability Mass Function of a Discrete Random Variable.


📘 Probability Mass Function(PMF):
It gives the exact value of a probability for a discrete random variable at a specific value \(x\).
It assigns a “non-zero” mass or probability to a specific countable outcome.
Note: Called ‘mass’ because probability is concentrated at a single discrete point.
\(PMF = P(X=x)\)
e.g: Bernoulli, Binomial, Multinomial, Poisson etc.

Commonly visualised as a bar chart.
Note: PMF = Jump at a given point in CDF.

\(PMF = P_x(X=x_i) = F_X(X=x_i) - F_X(X=x_{i-1})\)


📘 Bernoulli Distribution:
It models a single event with two possible outcomes, success (1) or failure (0), with a fixed probability of success, ‘p’.
p = Probability of success
1-p = Probability of failure
Mean = p
Variance = p(1-p)
Note: A single trial that adheres to these conditions is called a Bernoulli trial.
\(PMF, P(x) = p^x(1-p)^{1-x}\), where \(x \in \{0,1\}\)
For example:

  1. Toss a coin, we get heads or tails.
  2. Result of a test, pass or fail.
  3. Machine learning, binary classification model.

📘

Binomial Distribution:
It extends the Bernoulli distribution by modeling the number of successes that occur over a fixed number of independent trials.
n = Number of trials
k = Number of successes
p = Probability of success
Mean = np
Variance = np(1-p)

\(PMF, P(x=k) = \binom{n}{k}p^k(1-p)^{n-k}\), where \(k \in \{0,1,2,3,...,n\}\)
\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) i.e number of ways to achieve ‘k’ successes in ‘n’ independent trials.

Note: Bernoulli is a special case of Binomial distribution where n = 1.
Also read about Multinomial distribution i.e where number of possible outcomes is > 2.

For example:

  • Counting number of heads(success) in ’n’ coin tosses.




💡 What is the probability of getting exactly 2 heads in 3 coin tosses?

Total number of outcomes in 3 coin tosses = 2^3 = 8
Desired outcomes i.e 2 heads in 3 coin tosses = \(\{HHT, HTH, THH\}\) = 3
Probability of getting exactly 2 heads in 3 coin tosses = \(\frac{3}{8}\) = 0.375
Now lets solve the question using the binomial distribution formula.

\[P(k=2) = \binom{3}{2}p^2(1-p)^{3-2} \\ = \frac{3!}{2!(3-2)!}(0.5)^2(0.5) \\ = 3*0.25*0.5 = 3*0.125 = 0.375\]



💡 What is the probability of winning a lottery 1 out of 10 times, given that the probability of winning a single lottery = 1/3?

Number of successes, k = 1
Number of trials, n = 10
Probability of success, p = 1/3
Probability of winning lottery, P(k=1) =

\[\binom{10}{1}p^1(1-p)^{10-1} \\ = \frac{10!}{1!(10-1)!}(1/3)^1(2/3)^9 \\ = 10*0.333*0.026 = 0.866 \approx 8.66\% \]




📘

Poisson Distribution:
It expresses the probability of an event happening a certain number of times ‘k’ within a fixed interval of time.
Given that:

  1. Events occur with a known constant average rate.
  2. Occurrence of an event is independent of the time since the last event.

    Parameters:
    \(\lambda\): Expected number of events per interval
    \(k\) = Number of events in the same interval
    Mean = \(\lambda\)
    Variance = \(\lambda\)

PMF = Probability of occurrence of ‘k’ events in the same interval

\[PMF = \lambda^ke^{-\lambda}/k!\]

Note: Useful to count data where total population size is large but the probability of an individual event is small.

For example:

  1. Model the number of customers arrival at a service center per hour.
  2. Number of website clicks in a given time period.

💡 A company receives, on an average, 5 customer emails per hour. What is the probability of receiving exactly 3 emails in the next hour?

Expected (average) number of emails per hour, \(\lambda\) = 5
Probability of receiving exactly k=3 emails in the next hour =

\[P(k=3) = \lambda^3e^{-\lambda} / 3! \\ = 5^3e^{-5} / 3! = 125*e^{-5} / 6 \\ = 125*0.00674 / 6 \approx 0.14 ~or~ 14\% \]





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