In this chapter we will learn to calculate probability of coin toss with different variations.

Solved examples are also provided for your better understanding.

## Understanding coin toss probability

Before calculating the probability, let us revise the basics first.

Formula for probability

Probability tells the **chance of happening any event.**

Formula for probability is given as;

\mathtt{Probability\ =\ \frac{Number\ of\ favorable\ outcome}{Total\ possible\ outcomes}}

The above formula is used extensively in probability theory, so make sure you memorize it.

### Calculating probability of toss of unbiased coin

Consider the **experiment of tossing an unbiased coin**.

In this case there are two possible outcomes; Head or Tails.

**The sample space is given as;**

S = { Head , Tail }

So total number of possible outcome = 2.**Now let’s calculate;**

(a) probability of getting a head

(b) Probability of getting a tail**(a) Probability of getting a head**

Let the event is expressed by A.

A = { Head }

Number of favorable outcome = 1

Using the probability formula;

Probability = 1 / 2

Hence, there is 1/2 change of getting a head.

To calculate the probability on percentage, multiply the number by 100.

⟹ 1/2 x 100

⟹ 50%

Hence, there is 50% chance of getting head after tossing of unbiased coin.

**(b) Probability of getting a tail**

Let the event be expressed by B.

B = { Tails }

Number of favorable outcome = 1

Probability = 1/2

Probability (in percentage) = 1/2 x 100 = 50%

Hence, there is 50% chance of getting tail in toss of coin.

### Questions on Probability of coin

**Example 01**

Two coins are tossed simultaneously. Find the probability of;

(a) getting 2 heads

(b) exactly one head

(c) exactly one tail

(d) no tails

**Solution**

Let us first find the **sample space of tossing two coins.**

S = { HH, HT, TH, TT }

So number of total outcomes = 4**(a) Getting two heads**

Let A be the event of getting two heads.

A = { HH }

Number of favorable outcome = 1

Probability P(A) = 1/4

P(A) (in percentage) = 1/4 x 100 = 25%

Hence, there is 25% change of getting two heads.**(b) Exactly one head**

Let B be the event of getting exactly one head.

B = { HT, TH }

Number of favorable outcome = 2

Probability P(B) = 2/4 = 1/2

P(B) (in percentage) = 1/2 x 100 = 50%

Hence, there is 50% chance of getting exactly one head.

**(c) Exactly one tail**

Let C be the event of getting exactly one tail.

C = { HT, TH }

Number of favorable outcome = 2

Probability P(C) = 2/4 = 1/2

P(C) (in percentage ) = 1/2 x 100 = 50%

Hence, there is 50% chance of getting exactly one tails.

**(d) No tails**

Let D be the event of getting no tails.

D = { HH }

Number of favorable outcome = 1

P(D) = 1/4

P(D) in percentage) = 1/4 x 100 = 25%

Hence, there is 25% chance of getting no tails.

**Example 02**

Three coins are tossed simultaneously. Calculate the probability of getting;

(a) atleast two heads

(b) exactly 2 tails

(c) a head on first coin

**Solution**

Let us first write the **sample space of given experiment**.

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

Number of possible outcome = 8

**(a) At least two heads**

Let A be the event of getting atleast two heads.

A = { HHH, HHT, HTH, THH }

Number of favorable outcomes = 4

Probability P(A) = 4/8 = 1/2

P(A) (in percentage) = 1/2 x 100 = 50%

**(b) Exactly two tails.**

Let B be the event of getting two tails.

B = ( HTT, TTH, THT }

Number of favorable outcome = 3

Probability P(B) = 3 / 8 = 3 / 8

P(B) (in percentage) = 3/8 x 100 = 37.5%

Hence, the probability of getting exactly two tails is 37.5%

**(c) a head on first coin**

Let C be the event of getting a head on first coin.

C = { HHH, HHT, HTH, HTT }

Number of favorable outcome = 4

P(C) = 4 / 8 = 1/2

P(C) (in percent) = 1/2 x 100 = 50%

Hence, there is 50% chance of getting head on first coin.

**Example 03**

A coin is tossed. If it’s heads, a die is thrown and if it’s a tail then the same coin is tossed again. Find the probability of;

(a) two tails

(b) head and odd number

**Solution**

Let us write the **sample space** for this experiment;

S = { (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, H), (T, T) }

Number of total outcome = 8

**(a) getting two heads**

Let A be the event of getting two heads;

A = { (T, T) }

Number of possible outcome = 1

Probability P(A) = 1/8

P(A) (in percentage) = 1/8 x 100 = 12.5 %

Hence, there is 12.5% chance of getting two heads.

**(b) head and odd number**

Let B be the event of getting head and odd number.

B = { (H, 1), (H, 3), (H, 5) }

Number of favorable outcome = 3

Probability P(B) = 3/8

Probability P(B) in percentage = 3/8 x 100 = 37.5%

Hence, there is 37.5% chance of getting head and odd number