Two dice are thrown together and the total score is noted. The events E, F, and G are 'a total of 4', 'a total of 9 or more', and 'a total divisible by 5', respectively.

Calculate $P\left(E\right)$ , $P\left(F\right)$ , and $P\left(G\right)$ and decide which pairs of events, if any, are independent.

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Twodicearethrowntogether\\ \therefore n\left(S\right)=36\\ E=Atotalof4=\left\{\left(2,2\right),\left(1,3\right),\left(3,1\right)\right\}\therefore n\left(E\right)=3\\ F=Atotalof9ormore\\ =\left\{\left(3,6\right),\left(6,3\right),\left(5,4\right),\left(4,5\right),\left(5,5\right),\left(4,6\right),\left(6,4\right),\left(5,6\right),\left(6,5\right),\left(6,6\right)\right\}\therefore n\left(F\right)=10\\ G=Atotaldivisibleby5\\ =\left\{\left(1,4\right),\left(4,1\right),\left(2,3\right),\left(3,2\right),\left(4,6\right),\left(6,4\right),\left(5,5\right)\right\}\therefore n\left(G\right)=7\\ Here,weseethat\left(E\cap F\right)=?and\left(E\cap G\right)=?\\ and\left(F\cap G\right)=\left\{\left(4,6\right),\left(6,4\right),\left(5,5\right)\right\}\\ \therefore n\left(F\cap G\right)=3and\left(E\cap F\cap G\right)=?\\ \therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{36}=\frac{1}{12}\\ P\left(F\right)=\frac{n\left(F\right)}{n\left(S\right)}=\frac{10}{36}=\frac{5}{18}andP\left(G\right)=\frac{n\left(G\right)}{n\left(S\right)}=\frac{7}{36}\\ P\left(F\cap G\right)=\frac{3}{36}=\frac{1}{12}andP\left(F\right).P\left(G\right)=\frac{5}{18}.\frac{7}{36}=\frac{35}{648}\\ Since,P\left(F\cap G\right)\ne P\left(F\right).P\left(G\right)\\ Hence,thereisnopairofindependentevents.\end{array}$