Probability — Class 12 Maths Solution

.: If a fair dice is rolled, then the sample space is

$S = \{ 1,2,3,4,5,6\}$

and given that $E = \{ 1,3,5\} ,F = \{ 2,3\}$

and $G = \{ 2,3,4,5\}$

Hence $E \cap F = F \cap E = \{ 3\} ,E \cap G = G \cap E = \{ 3,5\} ,E \cup F = \{ 1,2,3,5\}$

$(E \cup F) \cap G = \{ 2,3,5\} ,(E \cap F) \cap G = \{ 3\}$

$P(E) = 3/6,P(F) = 2/6,P(G) = 4/6,P(E \cap F) = 1/6,P(E \cap G) = 2/6$

$P(E \cup F) = 4/6,P((E \cup F) \cap G) = 3/6,P((E \cap F) \cap G) = 1/6$

(i) $P(E|F) = \cfrac{{P(E \cap F)}}{{P(F)}} = \cfrac{{1/6}}{{2/6}} = \cfrac{1}{2}$

and
$P(F|E) = \cfrac{{P(E \cap F)}}{{P(E)}} = \cfrac{{1/6}}{{3/6}} = \cfrac{1}{3}$

(ii) $P(E|G) = \cfrac{{P(E \cap G)}}{{P(G)}} = \cfrac{{2/6}}{{4/6}} = \cfrac{2}{4} = \cfrac{1}{2}$

and $P(G|E) = \cfrac{{P(G \cap E)}}{{P(E)}} = \cfrac{{2/6}}{{3/6}} = \cfrac{2}{3}$

(iii) $P((E \cup F)|G) = \cfrac{{P((E \cup F) \cap G)}}{{P(G)}} = \cfrac{{3/6}}{{4/6}} = \cfrac{3}{4}$

and $P((E \cap F)|G) = \cfrac{{P((E \cap F) \cap G)}}{{P(G)}} = \cfrac{{1/6}}{{4/6}} = \cfrac{1}{4}$