Processing math: 100%

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: 

(i) getting three heads,

(ii) getting two heads,

(iii) getting one head,

(iv) getting no head

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,

(i) getting three heads

P(getting three heads) = P(E1)

      Number of times three heads appeared
=                   Total number of trials         

= 70/250

= 0.28


(ii) getting two heads

P(getting two heads) = P(E2)

      Number of times two heads appeared
=                 Total number of trials         

= 55/250

= 0.22


(iii) getting one head

P(getting one head) = P(E3)

      Number of times one head appeared
=                 Total number of trials        

= 75/250

= 0.30


(iv) getting no head

P(getting no head) = P(E4)

      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

Probability of Tossing Three Coins

2. When 3 unbiased coins are tossed once.

What is the probability of:

(i) getting all heads

(ii) getting two heads

(iii) getting one head

(iv) getting at least 1 head

(v) getting at least 2 heads

(vi) getting atmost 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.

(i) getting all heads

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.

(ii) getting two heads

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.

(iii) getting one head

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.

(vi) getting atmost 2 heads

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.


Outcomes

3 heads

2 heads

1 head

 No head

Total

Frequencies

48

64

100

38

250


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

(i) 1 head

(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

                                                   = Frequency of Favourable TrialsTotal Number of Trials

                                                   = Number of Times 1 Head AppearsTotal Number of Trials

                                                   = 100250

                                                   = 25


(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

                                         = Number of Times 2 Heads and 1 Trial appearsTotal Number of Trials

                                         = 64250

                                         = 32125


(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

                                                   = Number of Times No Head AppearsTotal Number of Trials

                                                   = 38250

                                                   = 19125.

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

Probability

Probability

Random Experiments

Experimental Probability

Events in 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

Probability for Rolling Two Dice

Solved Probability Problems

Probability for Rolling Three Dice






9th Grade Math

From Probability of Tossing Three Coins to HOME PAGE




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.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Multiplication of Decimal Numbers | Multiplying Decimals | Decimals

    May 03, 25 04:38 PM

    Multiplication of Decimal Numbers
    The rules of multiplying decimals are: (i) Take the two numbers as whole numbers (remove the decimal) and multiply. (ii) In the product, place the decimal point after leaving digits equal to the total…

    Read More

  2. Magic Square | Add upto 15 | Add upto 27 | Fibonacci Sequence | Videos

    May 03, 25 10:50 AM

    check the magic square
    In a magic square, every row, column and each of the diagonals add up to the same total. Here is a magic square. The numbers 1 to 9 are placed in the small squares in such a way that no number is repe

    Read More

  3. Division by 10 and 100 and 1000 |Division Process|Facts about Division

    May 03, 25 10:41 AM

    Divide 868 by 10
    Division by 10 and 100 and 1000 are explained here step by step. when we divide a number by 10, the digit at ones place of the given number becomes the remainder and the digits at the remaining places…

    Read More

  4. Multiplication by Ten, Hundred and Thousand |Multiply by 10, 100 &1000

    May 01, 25 11:57 PM

    Multiply by 10
    To multiply a number by 10, 100, or 1000 we need to count the number of zeroes in the multiplier and write the same number of zeroes to the right of the multiplicand. Rules for the multiplication by 1…

    Read More

  5. Adding and Subtracting Large Decimals | Examples | Worksheet | Answers

    May 01, 25 03:01 PM

    Here we will learn adding and subtracting large decimals. We have already learnt how to add and subtract smaller decimals. Now we will consider some examples involving larger decimals.

    Read More