Question

Two coins are tossed simultaneously. Find the probability of getting: (i)Two heads, (ii)At least one head, (iii)No head

Answer 1

Step 1: Use the formula of probability

$P\left(A\right)=\frac{numberofoutcomefavorabletoA}{totalnumberofPossibleoutcome}$

or

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

• $P\left(A\right)$ is probability of an event$‘A’$
• $n\left(A\right)$ is the favorable outcome of the event $‘A’$
• $n\left(S\right)$ is total number of possible outcomes in a sample space

Step 2: Find the probability of Two heads

Let A be the event of “getting two heads”

$S=\{HH,HT,TH,TT\}$

$S$ is a sample space of all the possible outcomes which is $4$

Thus, total number of possible outcomes in a sample space is

$n\left(S\right)=4$

In these 4 outcomes only one contain exactly two head

$A=\left\{HH\right\}$

Thus, the favorable outcome of the event $‘A’$ is $n\left(A\right)=1$

The probability of two head $P\left(A\right)$ is,

$$\begin{gathered} P\left(A\right)=\frac{numberoftwohead}{totalnumberofpossibleoutcomes} \\ =\frac{n\left(A\right)}{n\left(S\right)} \end{gathered}$$

Substitute the values on the formula,

$requiredprobability=\frac{1}{4}$

Step 3:Find the probability of at least one head

Let A be the event of "getting at least one head"

$S=\{HH,HT,TH,TT\}$

$S$ is a sample space of all the possible outcomes which is $4$

Thus, total number of possible outcomes in a sample space is

$n\left(S\right)=4$

In these 4 outcomes only three contain at least one head

$A=\{HH,HT,TH\}$

Thus, the favorable outcome of the event $‘A’$ is $n\left(A\right)=3$

The probability of at least one head $P\left(A\right)$ is,

$$\begin{gathered} P\left(A\right)=\frac{numberofatleastonehead}{totalnumberofpossibleoutcomes} \\ =\frac{n\left(A\right)}{n\left(S\right)} \end{gathered}$$

Substitute the values on the formula,

$requiredprobability=\frac{3}{4}$

Step 4: Find the probability of no head

Let A be the event of "getting no head"

$S=\{HH,HT,TH,TT\}$

$S$ is a sample space of all the possible outcomes which is $4$

Thus, total number of possible outcomes in a sample space is

$n\left(S\right)=4$

In these 4 outcomes only one contain no head

$A=\left\{TT\right\}$

Thus, the favorable outcome of the event $‘A’$ is $n\left(A\right)=1$

The probability of no head $P\left(A\right)$ is,

$$\begin{gathered} P\left(A\right)=\frac{numberofnohead}{totalnumberofpossibleoutcomes} \\ =\frac{n\left(A\right)}{n\left(S\right)} \end{gathered}$$

Substitute the values on the formula,

$requiredprobability=\frac{1}{4}$

Hence, the probability of getting two heads is $\frac{1}{4}$,the probability of At least one head is $\frac{3}{4}$,the probability of getting no head is $\frac{1}{4}$.