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