[SOLVED] AI5002-Assignment 4

Description

. Problem

Find the probability distribution of

(i) number of heads in two tosses of a coin. (ii) number of tails in the simultaneous tosses of three coins.

(iii) number of heads in four tosses of a coin.

1. Solution

Let Y denote the event of tossing a coin. Considering a fair coin, the probability of getting a Head

or Tail P(Y) = 0.5

We have i coin tosses, with probability p of Heads and (1-p) of Tails. We conduct the trials independently.

In general , the probability of getting of j Head/Tail is given as:

n!         _j                    (n−j)

P(Y = j) = p (1 − p) (1) j!(n − j)!

Consider a random variable X where X = Number of successes.Suppose we have n trials. We write X

˜B(n, p)

(i) Let X denote the random variable of number of Heads. The probability distribution of getting exactly j Heads in 2 tosses of coin is given as:

X ˜B(2, 0.5)

Using equation

2!         .5j(1 − 0.5)(2−j)             (2)

P(X = j) = 0 j!(2 − j)!

We get the pdf as below:

P(X = 0) = .5⁰(1 − 0.5)⁽²⁻⁰⁾ = 0.25 (3)

P(X = 1) = 0.5¹(1 − 0.5)⁽²⁻¹⁾ = 0.5 (4)

P(X = 2) = .5²(1 − 0.5)⁽²⁻²⁾ = 0.25 (5)

The distribution table is given as:

j	0	1	2
P(X=j)	0.25	0.5	0.25

(ii)Let X denote the random variable of number of Tails. The probability distribution of getting exactly j Tails in 3 tosses of coin is given as:

X ˜B(3, 0.5)

Using equation

3!         .5j(1 − 0.5)(3−j)             (6)

P(X = j) = 0 j!(3 − j)!

We get the pdf as below:

P(X = 0) = 0.5⁰(1 − 0.5)⁽³⁻⁰⁾ = 0.125

(7)

P(X = 1) = 0.5¹(1 − 0.5)⁽³⁻¹⁾ = 0.375

(8)

P(X = 2) = 0.5²(1 − 0.5)⁽³⁻²⁾ = 0.375

(9)

P(X = 3) = 0.5³(1 − 0.5)⁽³⁻³⁾ = 0.125

(10)

The probability distribution of X is:

j	0	1	2	3
P(X=j)	0.125	0.375	0.375	0.125

(iii)Let X denote the random variable of number of Heads. The probability distribution of getting exactly j Heads in 4 tosses of coin is given as:

X ˜B(4, 0.5)

Using equation

4!            ^{j                        (4−j)}

P(X = j) = 0.5 (1 − 0.5) (11) j!(4 − j)!

 

2

We get the pdf as below:

P(Y3 = 0) = 0.5⁰(1 − 0.5)⁽⁴⁻⁰⁾ = 0.0625

(12)

P(X = 1) = 0.5¹(1 − 0.5)⁽⁴⁻¹⁾ = 0.25

(13)

P(X = 2) = 0.5²(1 − 0.5)⁽⁴⁻²⁾ = 0.375

(14)

P(X = 3) = 0.5³(1 − 0.5)⁽⁴⁻⁴⁾ = 0.25

(15)

P(X = 4) = 0.5⁴(1 − 0.5)⁽⁴⁻⁴⁾ = 0.0625

(16)

The probability distribution of X is:

j	0	1	2	3	4
P(X=j)	0.0625	0.25	0.375	0.25	0.0625

The probabilities were simulated using the python code.

Figure 1: Simulation for tossing a fair coin