![\[(2n+1)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-78b5bf57e27382de024d05e8d0d3a07f_l3.png "Rendered by QuickLaTeX.com")
![\[31/42\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-849c6b533cd29d9a5cbe8eee0858947f_l3.png "Rendered by QuickLaTeX.com")
Given, n coins are two headed coins and the remaining
![\[(n+1)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4d92d8c768fd3b8f08e38ac7b18836b6_l3.png "Rendered by QuickLaTeX.com")
coins are fair.
Let
![\[{{E}_{1}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a18b4f4210098684c3183f0c3d29ae73_l3.png "Rendered by QuickLaTeX.com")
: the event that unfair coin is selected
![\[{{E}_{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0730822ca38b76c410dc8658f334d0de_l3.png "Rendered by QuickLaTeX.com")
: the event that the fair coin is selected
E: the event that the toss results in a head
So,
![\[P({{E}_{1}})\text{ }=\text{ }n/\left( 2n\text{ }+\text{ }1 \right)\text{ }and\text{ }P({{E}_{2}})\text{ }=\text{ }\left( n\text{ }+\text{ }1 \right)/\text{ }\left( 2n\text{ }+1 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9e952a971e328f4d17387b9041745701_l3.png "Rendered by QuickLaTeX.com")
![\[P(E/{{E}_{1}})\text{ }=\text{ }1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0949e03f228e5979fb7ebc41f2734868_l3.png "Rendered by QuickLaTeX.com")
(As it’s a sure event)
![\[P(E/{{E}_{2}})\text{ }=\text{ 1/2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a58bf761a10dacfc053da7c90f2377dd_l3.png "Rendered by QuickLaTeX.com")
Therefore, the required value of n is
![\[10\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7e13914576c8f07fa7780b4b1aae684d_l3.png "Rendered by QuickLaTeX.com")
.