Bernoulli Distributions

Bernoulli Distribution

If you understand Discrete Random Variables, then you'll find bernoulli distributions a breeze - because a bernoulli distribution is a form of discrete random variable. What makes a distribution a Bernoulli distribution is if it has only two outcomes (only two values for \(x\)).

One outcome typically represents 'success', whilst the other represents 'failure'. This can be used to model situations involving two possible outcomes. E.g. whether a head faces upwards from a coin toss.

Binomial Distribution

If we conduct a Bernoulli distribution many times, we create a Binomial distribution. Each case of Bernoulli Distribution conducted is named a 'trial' for the resulting Binomial Distribution.

A Binomial distribution has two definining properties:

1. \(n\): The number of trials
2. \(p\): The probability of success

Binomial Distribution is notated with:

\(X \sim Bin(n, p)\)

Mean and Variance

Here are some formulas for Binomial Distribution

\(\text{1)} \ X \sim Bin(n, p) \text{. The standard deviation is 4.} \ E(X) = 24 \text{. Find } n, p\).

\(\displaystyle \begin{cases} 4 = \sqrt{np(1-p)} & \\ np = 24 \end{cases} \)

Solve simultaneoulsy using ClassPad:

\(n=4, \ \ p = 0.3333 \)

In the game CS:GO, players can gamble on cases and recieve skin drops that hold real monetary value and can be sold. The rarest type of item is a knife, which has an estimated drop rate of 0.1% per case. What is the probability as a percent, that a player will consecutively open two knifes given they unlock 10 cases?

Let \(X \sim Bin(2, 0.001)\)

\(\begin{aligned} P(X=2) &= ^{n}\textrm{C}_x p^x (1-p)^{n-x} \\[5pt] &= ^{10}\textrm{C}_2 0.001^2 (1-0.001)^{10-2} \\[5pt] &= 45 \times 0.00001(0.999)^{3} \\[5pt] &= 0.0000448651 \\[5pt] &= 0.004487 \% \end{aligned} \)

By the way, I should mention that I play CS:GO, and on my first case I opened a butterfly knife (black laminate), well-worn, that was values at $1500 AUD. I have never opened a case since, making my knife drop rate 100%. :)