Question

Three coins are tossed together. Find the probability of getting(i)exactly two heads(ii)At least two heads(iii) At least one head and one tail

Answer 1

Step 1: Use the formula of probability

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

or

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

where,

• $P\left(A\right)$ 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 getting exactly two heads

Let $A$ be the event of "getting exactly two heads"

$S=\{HHH,HHT,HTH,THH,TTH,THT,HTT,TTT\}$

$S$ is a sample space of all the possible outcomes which is $8$

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

$n\left(S\right)=8$

In these $8$ outcomes only three contain exactly two heads

Getting exactly two head $A=\{HHT,HTH,THH\}$

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

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

$P\left(A\right)=\frac{numberofexactlytwohead}{totalnumberofpossibleoutcomes}$

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

Substitute the values on the formula,

$requiredprobability=\frac{3}{8}$

Step 3:Find the probability of getting at least two head

Let $A$ be the event of "getting at least two head"

$S=\{HHH,HHT,HTH,THH,TTH,THT,HTT,TTT\}$

$S$ is a sample space of all the possible outcomes which is $8$

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

$n\left(S\right)=8$

In these $8$outcomes only four contain at least two heads

Getting at least two head $A=\{HHT,HTH,THH,HHH\}$

Thus, the favorable outcome of the event ‘$A$’ is $n\left(A\right)=4$

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

$P\left(A\right)=\frac{numberofatleasttwohead}{totalnumberofpossibleoutcomes}$

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

Substitute the values on the formula,

$requiredprobability=\frac{4}{8}=\frac{1}{2}$

Step 4: Find the probability of getting at least one head and one tail

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

$S=\{HHH,HHT,HTH,THH,TTH,THT,HTT,TTT\}$

$S$ is a sample space of all the possible outcomes which is $8$

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

$n\left(S\right)=8$

In these $8$ outcomes only six contain at least one head and one tail

Getting at least one head and one tail

$A=\{HHT,HTH,THH,TTH,THT,HTT\}$

Thus, the favorable outcome of the event ‘$A$’ is $n\left(A\right)=6$

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

$P\left(A\right)=\frac{numberofatleastoneheadandonetail}{totalnumberofpossibleoutcomes}$

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

Substitute the values on the formula,

$requiredprobability=\frac{6}{8}=\frac{3}{4}$

Hence, the probability of getting two heads $\frac{3}{8}$, the probability of at least two head $\frac{1}{2}$, The probability of getting at least one head and one tail $\frac{3}{4}$.