# Coin Toss Probability

Problems on coin toss probability are explained here with different examples.

When we flip a coin there is always a probability to get a head or a tail is 50 percent.

Suppose a coin tossed then we get two possible outcomes either a ‘head’ (H) or a ‘tail’ (T), and it is impossible to predict whether the result of a toss will be a ‘head’ or ‘tail’.

The probability for equally likely outcomes in an event is:

Number of favourable outcomes ÷ Total number of possible outcomes

Total number of possible outcomes = 2

(i) If the favourable outcome is head (H).

Number of favourable outcomes = 1.

Number of favorable outcomes
= P(H) =
total number of possible outcomes

= 1/2.

(ii) If the favourable outcome is tail (T).

Number of favourable outcomes = 1.

Therefore, P(getting a tail)

Number of favorable outcomes
= P(T) =
total number of possible outcomes

= 1/2.

Word Problems on Coin Toss Probability:

1. A coin is tossed twice at random. What is the probability of getting

(ii) the same face?

Solution:

The possible outcomes are HH, HT, TH, TT.

So, total number of outcomes = 4.

(i) Number of favourable outcomes for event E

= Number of outcomes having at least one head

= 3 (as HH, HT, TH are having at least one head).

So, by definition, P(F) = 34

34.

(ii) Number of favourable outcomes for event E

= Number of outcomes having the same face

= 2 (as HH, TT are have the same face).

So, by definition, P(F) = 24

24 = 12

12.
2. If three fair coins are tossed randomly 175 times and it is found that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and zero head appeared 35 times.

What is the probability of getting

Solution:

Total number of trials = 175.

Number of times three heads appeared = 21.

Number of times two heads appeared = 56.

Number of times one head appeared = 63.

Number of times zero head appeared = 35.

(i)

Number of times three heads appeared
= P(E1) =
total number of trials

= 21/175

= 0.12

Number of times two heads appeared
= P(E2) =
total number of trials

= 56/175

=
0.32

Number of times one head appeared
= P(E3) =
total number of trials

=
63/175

= 0.36

Number of times zero head appeared
= P(E4) =
total number of trials

= 35/175

= 0.20

Note: Remember when 3 coins
are tossed randomly, the only possible outcomes

are E2, E3, E4 and

P(E1) + P(E2) + P(E3) + P(E4)

= (0.12 + 0.32 + 0.36 + 0.20)

= 1

3. Two coins are tossed randomly 120 times and it is found that two tails
appeared 60 times, one tail
appeared 48 times and no tail appeared 12 times.

If two coins are tossed at random, what is the
probability of getting

(i) 2 tails,

(ii) 1 tail,

(iii) 0 tail

Solution:

Total number of
trials = 120

Number of times 2 tails appear
= 60

Number of times 1 tail appears
= 48

Number of times 0 tail appears
=
12

Let E1, E2 and E3 be the events of getting 2 tails, 1 tail and 0 tail respectively.

(i) P(getting
2 tails)

Number of times 2 tails appear
= P(E1) =
total number of trials

= 60/120

= 0.50

(ii) P(getting 1 tail)

Number of times 1 tail appear
= P(E2) =
total number of trials

= 48/120

= 0.40

(iii) P(getting
0 tail)

Number of times no tail appear
= P(E3) =
total number of trials

= 12/120

= 0.10

Note:

Remember while tossing 2 coins simultaneously, the only possible outcomes are E1, E2, E3 and,

P(E1) + P(E2) + P(E3)

= (0.50 + 0.40 + 0.10)

= 1

4. Suppose a fair coin is randomly
tossed for 75 times and it is found that head turns up 45
times and tail 30 times. What is the probability of getting (i)
a head and (ii) a tail?

Solution:

Total number of trials = 75.

Number of times head turns up = 45

Number of times tail turns up = 30

(i) Let X be the event of

Number of times head turns up
= P(X) =
total number of trials

= 45/75

= 0.60

(ii) Let Y be
the event of getting a tail.

P(getting a tail)

Number of times tail turns up
= P(Y) =
total number of trials

= 30/75

= 0.40

Note: Remember when a
fair coin is tossed and then X and Y are
the only possible outcomes, and

P(X) + P(Y)

= (0.60 + 0.40)

= 1