What are the possible values of the random variable for the number of heads in tossing three coins?

First, let's think about what the probability of tossing a head or tail on one coin is. That would be 0.5 since the coin is "fair."

How about tossing three heads successively (HHH)? That would be 0.5 x 0.5 x 0.5, or (0.5)3

How about tossing two heads and then one tail in that order (HHT)? That would be 0.5 x 0.5 x 0.5, or (0.5)3

How about tossing two tails and then one head in that order (TTH)? That would again be (0.5)3.

Doing more of these examples, you can convince yourself that the probability of getting a particular outcome tossing three coins in a row is (0.5)3.

Let us now write out all the possible outcomes from tossing 3 coins in a row:

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

In other words, there are 8 possible outcomes, and each outcome has a probability of (0.5)3

If X represents the number of heads that will result from the 3 tosses, X can be 0, 1, 2, or 3 based on the above outcome (that is, 0 heads can result, 1 head can result, 2 head can result or 3 head can result).

Now let's consider these questions:

What is the probability that X = 0? That would correspond to the outcome TTT, which would be (0.5)3

What is the probability that X = 1? That would correspond to the outcomes HTT, THT, or TTH, depending on when the head comes. That would be (0.5)3 + (0.5)3 + (0.5)3 or 3(0.5)3

A probability distribution is basically all the possible values of the random variable and their corresponding probabilities. So for our case, this would be the probability distribution:

X = 0, (0.5)3

X = 1, 3(0.5)3

X = 2, 3(0.5)3

X = 3, (0.5)3

The sum of all probabilities in a probability distribution should be 1, which is true in our case.

Graphing this would be simple. Since this is a discrete function, you should have to denote each value of x and its probability using vertical lines.

E(X) is basically the expectant value, or weighted average, of the probability distribution. To calculate the E(X), we need to sum x*p(x) for all values of x. So in this case, it would be 0*(0.5)3 + 1*3(0.5)3 + 2*3(0.5)3 + 3*(0.5)3, which is 1.5

V(X) is the variance and is a measure of how far each random variable is from the mean.

To calculate the variance we need to average the square of the difference of each random variable and the mean. So, in other words, V(X) = E[(X-E(X))2]

So, (0-1.5)2 (0.5)3 + (1-1.5)2 3(0.5)3 + (2-1.5)2 3(0.5)3 + (3 -1.5)2 (0.5)3 = 0.75

Then we divide 5 by the number of trials, which in this case was 3 (since we tossed the coin 3 times). So 5/3 is the variance

Note: this is an example of the binomial distribution! You can read about it further online. Knowing that it is a binomial distribution can provide many useful shortcuts, like E(X) = np, where n = 3 and p = 0.5, or V(X) = np(1-p)

In this article, we will learn how to find the probability of tossing 3 coins. We know that when a coin is tossed, the outcomes are head or tail. We can represent head by H and tail by T. Now consider an experiment of tossing three coins simultaneously. The possible outcomes will be HHH, TTT, HTT, THT, TTH, THH, HTH, HHT. So the total number of outcomes is 23 = 8. The above explanation will help us to solve the problems of finding the probability of tossing three coins.

The probability of an event E is defined as P(E) = (Number of favourable outcomes of E)/ (total number of possible outcomes of E).

Step by Step Solutions to the tossing of 3 coins Problems

The following are some problems related to the tossing of 3 coins.

Example 1. When 3 unbiased coins are tossed once.

Find the probability of:

(i) getting all tails

(ii) getting two heads

(iii) getting at least 1 head

(iv) getting one head

Solution:

When 3 coins are tossed, the possible outcomes are HHH, TTT, HTT, THT, TTH, THH, HTH, HHT.

The sample space is S = { HHH, TTT, HTT, THT, TTH, THH, HTH, HHT}

Number of elements in sample space, n(S) = 8

(i) Let E1 denotes the event of getting all tails.

E1 = {TTT}

n(E1) = 1

P(getting all tails) = n(E1)/ n(S)

= ⅛

Hence the required probability is ⅛.

(ii) Let E2 denotes the event of getting two heads.

E2 = {HHT, HTH, THH}

n(E2) = 3

P(getting two heads) = n(E2)/ n(S)

= 3/8

Hence the required probability is ⅜.

(iii) Let E3 denotes the event of getting atleast one head.

E3 = { HHH, HTT, THT, TTH, THH, HTH, HHT }

n(E3) = 7

P(getting atleast one head) = n(E3)/ n(S)

= 7/8

Hence the required probability is 7/8.

(iv) Let E4 denotes the event of getting one head.

E4 = { HTT, THT, TTH}

n(E4) = 3

P(getting one head) = n(E4)/ n(S)

= 3/8

Hence the required probability is 3/8.

Example 2: In an experiment, three coins are tossed simultaneously at random 250 times. It was found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times. find the probability of:

(i) getting three heads,

(ii) getting one head,

(iii) getting no head

(iv) getting two heads,

Solution:

The total number of trials, n(S) = 250

(i) Let E1 denotes the event of getting three heads.

n(E1) = 70

P(E1) = No. of times 3 heads appeared/ total number of trials

= n(E1) / n(S)

= 70/250

= 0.28

Hence the required probability is 0.28.

(ii) Let E2 denotes the event of getting one head.

n(E2) = 75

P(E2) = No. of times 1 head appeared/ total number of trials

= n(E2) / n(S)

= 75/250

= 0.3

Hence the required probability is 0.3.

(iii) Let E3 denote the event of getting no head.

n(E3) = 50

P(E3) = No. of times no head appeared/ total number of trials

= n(E3) / n(S)

= 50/250

= 0.2

Hence the required probability is 0.2.

(iv) Let E4 denote the event of getting 2 heads.

n(E4) = 55

P(E4) = No. of times 2 heads appeared/ total number of trials

= n(E4) / n(S)

= 55/250

= 0.22

Hence the required probability is 0.22.

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The possible outcomes are HTT, THT, TTH, THH, HTH, HHT, HHH, and TTT.

When 3 coins are tossed, the number of outcomes = 23 = 8.

When 3 coins are tossed, the favourable outcome is HHH.
So required probability = 1/8.

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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

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

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

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

= \(\frac{100}{250}\)

= \(\frac{2}{5}\)

(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.

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