Question

Two coins are tossed. What is the probability of coming up with two heads, if it is known that at least one head comes up?

Answer 1

Hint: We know that if two coins are tossed, the sample space is given by
${\text{S = }}\left\{ {\left( {{\text{H,H}}} \right){\text{,}}\,\left( {{\text{H,T}}} \right){\text{,}}\,\left( {{\text{T,H}}} \right){\text{,}}\,\left( {{\text{T,T}}} \right)} \right\}$
Note that, at least one head can come up in three cases, while two heads can come up in only one case.
Therefore, find the conditional probability of coming up of two heads, if it is known that at least one head comes up.

Complete step-by-step answer:
We know that if two coins are tossed, the sample space is given by
${\text{S = }}\left\{ {\left( {{\text{H,H}}} \right){\text{,}}\,\left( {{\text{H,T}}} \right){\text{,}}\,\left( {{\text{T,H}}} \right){\text{,}}\,\left( {{\text{T,T}}} \right)} \right\}$
 Now, let the event of coming up of two heads be named as event A.
Therefore, ${\text{A = }}\left\{ {\left( {{\text{H,H}}} \right)} \right\}$
Probability of an event happening\[{\text{ = p(E) = }}\dfrac{{{\text{Number of ways it can happen}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{n(E)}}{{n(S)}}\]
Hence, Probability of coming up of two heads is given by
${\text{p(A) = }}\dfrac{{{\text{n}}\left( {\text{A}} \right)}}{{{\text{n}}\left( S \right)}} = \dfrac{1}{4}$,
Again, let the event of at least one head coming up, be named as event B.
Therefore, ${\text{B = }}\left\{ {\left( {{\text{H,H}}} \right){\text{,}}\,\left( {{\text{H,T}}} \right){\text{,}}\,\left( {{\text{T,H}}} \right)} \right\}$
Hence, Probability of coming up of at least one head is given by,
${\text{p(B) = }}\dfrac{{{\text{n(B)}}}}{{{\text{n(S)}}}}{\text{ = }}\dfrac{{\text{3}}}{{\text{4}}}$
Now we have to find the probability of coming up of two heads, when it is known that at least one head comes up, i.e. we have to find ${\text{p}}\left( {{\text{A|B}}} \right)$.
We know, by the formula of conditional probability, ${\text{p(A|B) = }}\dfrac{{{\text{p}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{p}}\left( {\text{B}} \right)}}$
Since ${\text{p}}\left( {{\text{A}} \cap {\text{B}}} \right) = \dfrac{{{\text{n}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \dfrac{1}{4}$
$ \Rightarrow {\text{p(A|B) = }}\dfrac{{\left( {\dfrac{1}{4}} \right)}}{{\left( {\dfrac{3}{4}} \right)}}$ ,
On simplification we get,
$ \Rightarrow {\text{p(A|B) = }}\dfrac{1}{{{4}}} \times \dfrac{{{4}}}{3} = \dfrac{1}{3}$
Therefore, if two coins are tossed, the probability of coming up with two heads if it is known that at least one head comes up, is $\dfrac{1}{3}$.


Note: Note that, \[{\text{probability of an event happening = p(E) = }}\dfrac{{{\text{Number of ways it can happen}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{n(E)}}{{n(S)}}\]
Also, by the formula of conditional probability, ${\text{p(A|B) = }}\dfrac{{{\text{p}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{p}}\left( {\text{B}} \right)}}$