COIN TOSSING PROBABILITY

Coin tossing experiment always plays a key role in probability concept. Whenever we go through the stuff probability in statistics, we will definitely have examples with coin tossing.

Sample Space

When a coin is tossed, there are two possible outcomes.

They are 'Head' and 'Tail'.

So, the sample space S = {H, T}, n(s) = 2.

When two coins are tossed once,

total number of all possible outcomes = 2 x 2 = 4

So, the sample space S  =  {HH, TT, HT, TH}, n(s) = 4.

When three coins are tossed once,

total no. of all possible outcomes = 2 x 2 x 2 = 8.

So, the sample space is

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

n(s) = 8

In this way, we can get sample space when a coin or coins are tossed.

Note :

In coin toss experiment, we can get sample space through tree diagram also.

Summary :

One coin is tossed once :

Total number of possible outcomes = 2¹ = 2

Two coins are tossed once (or) One coin is tossed twice :

Total number of possible outcomes = 2² = 4

Three coins are tossed once (or) One coin is tossed thrice :

Total number of possible outcomes = 2³ = 8

Four coins are tossed once (or) One coin is tossed four times :

Total number of possible outcomes = 2⁴ = 16

In this way, when 'n' coins are tossed once (or) One coin is tossed 'n' number of times :

Total number of possible outcomes = 2ⁿ

Probability Formula

We can use the formula from classic definition to find probability in coin tossing experiments.

Let A be the event in a random experiment.

Then,

n(A) = Number of possible outcomes for the event A

n(S) = Number of all possible outcomes of the experiment

Here "S" stands for sample space which is the set contains all possible outcomes of the random experiment.

Then the above formula will become.

To have better understanding of the above formula, let us consider the following coin tossing experiment. 

A coin is tossed once. 

S = {H, T} and n(S) = 2

Let A be the event of getting head. 

A = {H} and n(A) = 1

Then,

P(A) = n(A)/n(S)

= 1/2

Solved Problems

Problem 1 :

A coin is tossed twice. What is the probability of getting head ?

Solution :

When a coin is tossed twice, total number of all possible outcomes :

= 2 x 2

= 4

Sample space :

S = {HH, TT, HT, TH}

n(S) = 4

Letting A be the event of getting exactly one head.

A = {HT, TH}

n(A) = 2

Required probability :

P(A) = n(A)/n(S)

= 2/4

= 0.50 or 50%

Problem 2 :

A coin is tossed three times. What is the probability of getting :

(i) 2 tails

(ii) at least 2 tails

Solution :

When a coin is tossed three times, first we need enumerate all the elementary events.

This can be done using 'Tree diagram' as shown below :

Hence the elementary events are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

That is, 

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

Thus the number of elementary events n(s) = 8.

(i) 2 tails :

Out of these 8 outcomes, 2 tails occur in three cases namely TTH, THT and HTT.

If we denote the occurrence of 2 tails by the event A and if assume that the coin as well as performer of the experiment is unbiased then this assumption ensures that all the eight elementary events are equally likely.

Required probability :

P(A) = n(A)/n(s)

= 3/8  

= 0.375 or 37.5%

(ii) at least 2 tails :

Let B denote occurrence of at least 2 tails i.e. 2 tails or 3 tails.

Since 2 tails occur in 3 cases and 3 tails occur in only 1 case, B occurs in 3 + 1 or 4 cases.

Required probability :

P(B) = 4/8

= 0.50 or 50%

Problem 3 :

Four coins  are tossed once. What is the probability of getting at least 2 heads ?

Solution :

When four coins are tossed once, total no. of all possible outcomes :

= 2 x 2 x 2 x 2

= 16

Sample space :

S = {HHHH, TTTT, HHHT, HHTH, HTHH, THHH, TTTH, TTHT, THTT, HTTT, HHTT, TTHH, HTHT, THTH, HTTH, THHT}

n(S) = 16

Let A be the event of getting at least two heads.

Then A has to include all the events in which there are two heads and more than two heads. 

A = {HHHH, HHHT, HHTH, HTHH, THHH, TTHH, HHTT, THTH, HTHT, THHT, HTTH}

n(A) = 11

Required probability :

P(A) = n(A)/n(S)

= 11/16

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