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