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.
Therefore, P(getting a head)

               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. 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
(i) three heads, (ii) two heads, (iii) one head, (iv) 0 head.

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.

Let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and zero head respectively.

(i) P(getting three heads)

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

= 21/175

= 0.12

(ii) P(getting two heads)

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

= 56/175

= 0.32

(iii) P(getting one head)

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

= 63/175

= 0.36

(iv) P(getting zero head)

            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


2. 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


3. 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 getting a head.

P(getting a head)

            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

9 Grade Math

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