A man tosses two different coins (one of

    \[\mathbf{Rs}\text{ }\mathbf{2}\]

and another of

    \[\mathbf{Rs}\text{ }\mathbf{5}\]

) simultaneously. What is the probability that he gets:

    \[\left( \mathbf{i} \right)\]

at least one head?

    \[\left( \mathbf{ii} \right)\]

at most one head?
A man tosses two different coins (one of

    \[\mathbf{Rs}\text{ }\mathbf{2}\]

and another of

    \[\mathbf{Rs}\text{ }\mathbf{5}\]

) simultaneously. What is the probability that he gets:

    \[\left( \mathbf{i} \right)\]

at least one head?

    \[\left( \mathbf{ii} \right)\]

at most one head?

Solution:

We know that,

When two coins are tossed simultaneously, the possible outcomes are

    \[\left\{ \left( H,\text{ }H \right),\text{ }\left( H,\text{ }T \right),\text{ }\left( T,\text{ }H \right),\text{ }\left( T,\text{ }T \right) \right\}\]

So

    \[,\text{ }n\left( S \right)\text{ }=\text{ }4\]

    \[\left( i \right)\]

The outcomes favourable to the event E, ‘at least one head’ are

    \[\left\{ \left( H,\text{ }H \right),\text{ }\left( H,\text{ }T \right),\text{ }\left( T,\text{ }H \right) \right\}\]

So, the number of outcomes favourable to E is

    \[3\text{ }=\text{ }n\left( E \right)\]

Hence,

    \[P\left( E \right)\text{ }=\text{ }n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }{\scriptscriptstyle 3\!/\!{ }_4}\]

    \[\left( ii \right)\]

The outcomes favourable to the event E, ‘at most one head’ are

    \[\left\{ \left( T,\text{ }H \right),\text{ }\left( H,\text{ }T \right),\text{ }\left( T,\text{ }T \right) \right\}\]

So, the number of outcomes favourable to E is

    \[3\text{ }=\text{ }n\left( E \right)\]

Hence,

    \[P\left( E \right)\text{ }=~n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }3/4\]