Lakshmi tosses two coins simultaneously the probability that she gets at most one head is

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`

Text Solution

Solution : Let us start with taking Total number of outcomes be=4{HH,TT,TH,HT}<br> Let E=event of getting at least one head<br> Number of favourable outcomes be=3{HT,HH,TH}<br> P(E)=Number of favourable outcomes /Total number of outcomes<br> =`3/4`

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