Question

If three coins are tossed, represent the sample space and the event of getting at least two heads also find the number of elements in them.

• A. $n\left(S\right)=8$ and $n\left(E\right)=4$
• B. $n\left(S\right)=8$ and $n\left(E\right)=1$
• C. $n\left(S\right)=8$ and $n\left(E\right)=7$
• D. $n\left(S\right)=8$ and $n\left(E\right)=3$

Answer 1

The correct option is A $n\left(S\right)=8$ and $n\left(E\right)=4$

If 3 coins are tossed,then the possible sample space can be : $\left(HHH\right),\left(HHT\right),\left(HTH\right),\left(THH\right),\left(TTH\right),\left(THT\right),\left(HTT\right),\left(TTT\right)$
where $H$=heads,$T$=tails
Hence,$n\left(S\right)=8$ and $n\left(E\right)=4$.