# Probability of Tossing Three Coins

Here we will learn how to find the probability of tossing three coins.

Let us take the experiment of tossing three coins simultaneously:

When we toss three coins simultaneously then the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for head and T is denoted for tail.

Therefore, total numbers of outcome are 23 = 8

The above explanation will help us to solve the problems on finding the probability of tossing three coins.

Worked-out problems on probability involving tossing or throwing or flipping three coins:

1. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times.

If three coins are tossed simultaneously at random, find the probability of:

Solution:

Total number of trials = 250.

Number of times three heads appeared = 70.

Number of times two heads appeared = 55.

Number of times one head appeared = 75.

Number of times no head appeared = 50.

In a random toss of 3 coins, let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and 0 head respectively. Then,

Number of times three heads appeared
=                   Total number of trials

= 70/250

= 0.28

Number of times two heads appeared
=                 Total number of trials

= 55/250

= 0.22

Number of times one head appeared
=                 Total number of trials

= 75/250

= 0.30

Number of times on head appeared
=                 Total number of trials

= 50/250

= 0.20

Note:

In tossing 3 coins simultaneously, the only possible outcomes are E1, E2, E3, E4 and P(E1) + P(E2) + P(E3) + P(E4)

= (0.28 + 0.22 + 0.30 + 0.20)

= 1

2. When 3 unbiased coins are tossed once.

What is the probability of:

(iv) getting at least 1 head

(v) getting at least 2 heads

Solution:

In tossing three coins, the sample space is given by

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

And, therefore, n(S) = 8.

Let E1 = event of getting all heads. Then,
E1 = {HHH}
and, therefore, n(E1) = 1.
Therefore, P(getting all heads) = P(E1) = n(E1)/n(S) = 1/8.

Let E2 = event of getting 2 heads. Then,
E2 = {HHT, HTH, THH}
and, therefore, n(E2) = 3.
Therefore, P(getting 2 heads) = P(E2) = n(E2)/n(S) = 3/8.

Let E3 = event of getting 1 head. Then,
E3 = {HTT, THT, TTH} and, therefore,
n(E3) = 3.
Therefore, P(getting 1 head) = P(E3) = n(E3)/n(S) = 3/8.

(iv) getting at least 1 head

Let E4 = event of getting at least 1 head. Then,
E4 = {HTT, THT, TTH, HHT, HTH, THH, HHH}
and, therefore, n(E4) = 7.
Therefore, P(getting at least 1 head) = P(E4) = n(E4)/n(S) = 7/8.

(v) getting at least 2 heads

Let E5 = event of getting at least 2 heads. Then,
E5 = {HHT, HTH, THH, HHH}
and, therefore, n(E5) = 4.
Therefore, P(getting at least 2 heads) = P(E5) = n(E5)/n(S) = 4/8 = 1/2.

Let E6 = event of getting atmost 2 heads. Then,
E6 = {HHT, HTH, HTT, THH, THT, TTH, TTT}
and, therefore, n(E6) = 7.
Therefore, P(getting atmost 2 heads) = P(E6) = n(E6)/n(S) = 7/8

3. Three coins are tossed simultaneously 250 times and the outcomes are recorded as given below.

If the three coins are again tossed simultaneously at random, find the probability of getting

(ii) 2 heads and 1 tail

(iii) All tails

Solution:

(i) Total number of trials = 250.

Number of times 1 head appears = 100.

Therefore, the probability of getting 1 head

= $$\frac{\textrm{Frequency of Favourable Trials}}{\textrm{Total Number of Trials}}$$

= $$\frac{\textrm{Number of Times 1 Head Appears}}{\textrm{Total Number of Trials}}$$

= $$\frac{100}{250}$$

= $$\frac{2}{5}$$

(ii) Total number of trials = 250.

Number of times 2 heads and 1 tail appears = 64.

[Since, three coins are tossed. So, when there are 2 heads, there will be 1 tail also].

Therefore, the probability of getting 2 heads and 1 tail

= $$\frac{\textrm{Number of Times 2 Heads and 1 Trial appears}}{\textrm{Total Number of Trials}}$$

= $$\frac{64}{250}$$

= $$\frac{32}{125}$$

(iii) Total number of trials = 250.

Number of times all tails appear, that is, no head appears = 38.

Therefore, the probability of getting all tails

= $$\frac{\textrm{Number of Times No Head Appears}}{\textrm{Total Number of Trials}}$$

= $$\frac{38}{250}$$

= $$\frac{19}{125}$$.

These examples will help us to solve different types of problems based on probability of tossing three coins.

Probability

Probability

Random Experiments

Experimental Probability

Empirical Probability

Coin Toss Probability

Probability of Tossing Two Coins

Probability of Tossing Three Coins

Complimentary Events

Mutually Exclusive Events

Mutually Non-Exclusive Events

Conditional Probability

Theoretical Probability

Odds and Probability

Playing Cards Probability

Probability and Playing Cards

Solved Probability Problems

Probability for Rolling Three Dice

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

## Recent Articles

1. ### Adding 1-Digit Number | Understand the Concept one Digit Number

Sep 17, 24 02:25 AM

Understand the concept of adding 1-digit number with the help of objects as well as numbers.

2. ### Counting Before, After and Between Numbers up to 10 | Number Counting

Sep 17, 24 01:47 AM

Counting before, after and between numbers up to 10 improves the child’s counting skills.

3. ### Worksheet on Three-digit Numbers | Write the Missing Numbers | Pattern

Sep 17, 24 12:10 AM

Practice the questions given in worksheet on three-digit numbers. The questions are based on writing the missing number in the correct order, patterns, 3-digit number in words, number names in figures…

4. ### Arranging Numbers | Ascending Order | Descending Order |Compare Digits

Sep 16, 24 11:24 PM

We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order. Vice-versa while arranging numbers from the largest number to the sma…