Question

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Answer 1

Hint – In this question take every possible case that can constitute the overall possible outcome. One case can be all heads appearing in tossing of three coins, another case can be appearing of all tails while tossing, similarly there are two more cases. This approach will help get the answer.

Complete step-by-step answer:

As we know if a coin is tossed the possible number of outcomes is either head (h) or tail (t).
So when a coin is tossed three times so the set (S) of possible outcomes are:
Case (1) – when head appears all the time so the possible set is,
{hhh}
Case (2) – when head appears two time and tail appears one time so the possible set is,
{hht, hth, thh}.
Case (3) – when head appears one time and tail appears two time so the possible set is,
{htt, tht, tth}.
Case (4) – when tail appears all the time so the possible set is,
{ttt}.
So the required sample set or the total number of outcomes is
$ \Rightarrow S = \left\{ {hhh,hht,hth,thh,htt,tht,tth,ttt} \right\}$
So the number of possible outcomes are 8.
So this is the required answer.

Note – This concept serves as basics for solving probability problems, whenever coins are tossed or probability of any event is to be recorded then the overall possible cases has to be taken into consideration. This same concept is used even when we toss 2 coins or when the coins are increased from 3 to say 4.
There is a direct approach to solve these kind of problems as we know when a coin is tossed we will get either head or tail nothing else so there are two possible outcomes so when (n) number of coins are tossed the possible outcomes are $2 \times 2 \times 2 \times 2 \times ...........2 \times 2$ up to (n) times which is equal to ${2^n}$ so here there are 3 coins so, n = 3, so the possible number of outcomes are ${2^3} = 8$.