When two coins are tosses simultaneously , the possiuble outcomes are {(H,H)(H,T)(T,H)(T,T)} n(s)=4 The outcomes favourable to the event E,' at most one head are {(T,H)(H,T)(T,T)} So, the number of outcomes favourable to E is 3=n(E) Therefore, p(E)=`(n(E))/(n(s))=3/4` |