Probability of an event quantifies the chances of happening or non-happening of that event. The probability lies within 0 and 1. The problems related to toss of one or more coins are very common.

We know that an ideal coin is a circular disc with negligible thickness having two faces - head and tail. In the coin-toss problem, a coin is tossed in the air and the outcome is noted when it falls down.

The outcomes would be either a head (H) or a tails (T). The probability of obtaining both of them is equal.

For a fair coin, it is $\frac{1}{2}$. Also, When a coin is tossed, getting both head and tail at a time is an impossible event. Thus, its probability is zero.

It is a fact that the concept of probability does not tell what will happen, it merely indicates how likely an event is going to happen. Evidently, it must be very unlikely that on tossing twenty coins, they all result in heads.

But coin-toss probability is still very uncertain; it is possible that if lot many people toss a number of coins for a number of times, then this may happen. Also, if a coin is tossed 19 times and the head is obtained as a result each time, even then the next trial would have same odds of getting a tail, i.e. 50%. These tosses are all independent events.

There are several different kinds of problems related to the toss of a coin. In this article, we are going to go ahead and understand about this concept in detail and learn solved examples based on it.

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When a coin is tossed, we can get one of the two possible outcomes - head or tail. Both of them are equally likely. Similarly, on tossing two coins simultaneously, there may either two tails or two heads or head and tail. Roughly speaking, there would be three equally likely outcomes.

But actually this is not right. Since the two coins are separate, hence the order of getting a head and a tail also matters. In this way, there are four possible outcomes, such as -

head-head, head-tail, tail-head, tail-tail.

When more coins are tossed a number of times, there are more results and more probabilities as well. The number of possible outcomes in the toss of coin increased (in fact multiply) with the increase of number of coins. If total "n" number of coins are tossed, then there can be $2^{n}$ number of possible outcomes.

Let us suppose that the event of getting a head is denoted by "H" and that of tail is written by "T".

**The table of possible outcomes are listed below:**

When a single coin is tossed:

Number of possible outcomes = $2^{1}$ = 2

H, T

When two coins are tossed:

Number of possible outcomes = $2^{2}$ = 4

HT, TH, HH, TT

When three coins are tossed:

Number of possible outcomes = $2^{3}$ = 8

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

When four coins are tossed:

Number of possible outcomes = $2^{4}$ = 16

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT

When five coins are tossed:

Number of possible outcomes = $2^{5}$ = 32

HHHHH, HHHHT, HTTTH, HTTHT, HTTHH, HTHTT, HTHTH, HTHHT, HTHHH, HHTTT, HHTTH, HHTHT, HHTHH, HHHTT, HHHTH, HTTTT, TTTTT, TTTTH, TTTHT, TTTHH, TTHTT, TTHTH, TTHHT, TTHHH, THTTT, THTTH, THTHT, THTHH, THHTT, THHTH, THHHT, THHHH

For this process, list the possible outcomes and work out the probability by the above mentioned formula. The probabilities for different combinations of heads and tails can also be worked out by this method.
In order to calculate the probability problem involving the phrase "at least", one is required to list all the possible outcomes and count the favorable outcomes.

**For Example:** if a coin is tossed 3 times, then

The probability of getting at least two tails = probability of getting two tails + probability of getting three tails

Similarly,

On a toss of four coins,

The probability of getting at least two heads = probability of getting two heads + probability of getting three heads + probability of getting four heads

**The examples based on coin-toss probabilities are given below:**

**Example 1:** A coin is tossed 50 times getting 34 heads. What is the probability of tail.

**Solution:** Number of possible outcomes = 50

Number of favorable outcomes = 50 - 34 = 16

Probability of tail = $\frac{favorable\ outcomes}{Total\ outcomes}$

= $\frac{16}{50}$

= $\frac{8}{25}$ = 0.32

**Example 2:** Three coins are tossed. Find the following probabilities:

**(i)** getting all tails

**(ii)** getting two heads

**(iii)** getting at least one tail

**Solution:** On a toss of three coins, the possible outcomes would be:

**ii)** Number of favorable outcomes for two heads = 3 i.e. {HHT, HTH, THH}

**iii)** Number of favorable outcomes for at least one tail = 7 i.e. {HHT, HTH, THH, HTT, THT, TTH, TTT}

But actually this is not right. Since the two coins are separate, hence the order of getting a head and a tail also matters. In this way, there are four possible outcomes, such as -

head-head, head-tail, tail-head, tail-tail.

When more coins are tossed a number of times, there are more results and more probabilities as well. The number of possible outcomes in the toss of coin increased (in fact multiply) with the increase of number of coins. If total "n" number of coins are tossed, then there can be $2^{n}$ number of possible outcomes.

Let us suppose that the event of getting a head is denoted by "H" and that of tail is written by "T".

When a single coin is tossed:

Number of possible outcomes = $2^{1}$ = 2

H, T

When two coins are tossed:

Number of possible outcomes = $2^{2}$ = 4

HT, TH, HH, TT

When three coins are tossed:

Number of possible outcomes = $2^{3}$ = 8

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

When four coins are tossed:

Number of possible outcomes = $2^{4}$ = 16

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT

When five coins are tossed:

Number of possible outcomes = $2^{5}$ = 32

HHHHH, HHHHT, HTTTH, HTTHT, HTTHH, HTHTT, HTHTH, HTHHT, HTHHH, HHTTT, HHTTH, HHTHT, HHTHH, HHHTT, HHHTH, HTTTT, TTTTT, TTTTH, TTTHT, TTTHH, TTHTT, TTHTH, TTHHT, TTHHH, THTTT, THTTH, THTHT, THTHH, THHTT, THHTH, THHHT, THHHH

The answer of the question, "what is the probability of getting a head (or a tail) on a toss of a coin" is quite evidently "0.5 or $\frac{1}{2}$". But actually, this is the case when the coin is fair i.e. both events are equally likely.

This probability is calculated by the following formula:

P = $\frac{favorable\ outcomes}{Total\ outcomes}$

P = $\frac{favorable\ outcomes}{Total\ outcomes}$

Here, favorable outcomes = 1 (either head or tail as per question)

Total outcomes = 2 (head and tail)

Thus, P = $\frac{1}{2}$

But, if the coin is suspected to be not fair, then the coin should be tossed more than once (in fact a large number of times) and the heads (or tails) are counted. Let us suppose that on flipping a coin 100 times, we get 66 heads, its probability would be $\frac{66}{100}$ = 0.66.

This method of finding probability is called relative frequency or empirical probability. It has been noticed that as the number flips are increased, the probability becomes closer to $\frac{1}{2}$ or 0.5.

For this process, list the possible outcomes and work out the probability by the above mentioned formula. The probabilities for different combinations of heads and tails can also be worked out by this method.

The probability of getting at least two tails = probability of getting two tails + probability of getting three tails

Similarly,

On a toss of four coins,

The probability of getting at least two heads = probability of getting two heads + probability of getting three heads + probability of getting four heads

Number of favorable outcomes = 50 - 34 = 16

Probability of tail = $\frac{favorable\ outcomes}{Total\ outcomes}$

= $\frac{16}{50}$

= $\frac{8}{25}$ = 0.32

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

**i)** Number of favorable outcomes for getting all tails = 1 {TTT}

Thus, P(1) = $\frac{1}{8}$

Thus, P(1) = $\frac{1}{8}$

P(2) = $\frac{3}{8}$

P(3) = $\frac{7}{8}$

**Example 3:** A coin is flipped four times. What is the probability of obtaining at least two heads ?

**Solution:** On flipping a coin four times, the possible outcomes would be the following:

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT

Total outcomes = 16

Favorable outcomes of at least two heads are

{HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, THHH, TTHH, THTH, THHT}

i.e. 11.

P(at least two heads) = $\frac{11}{16}$ = 0.6875

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT

Total outcomes = 16

Favorable outcomes of at least two heads are

{HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, THHH, TTHH, THTH, THHT}

i.e. 11.

P(at least two heads) = $\frac{11}{16}$ = 0.6875

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