Probability of coin toss

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.

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;
(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

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.

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;
(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

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

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

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.

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