Solution plz

Let a_{1},a_{2},a_{3},a_{4},a_{5},a_{6} are the six faces

Probability of odd= 90% Probability of even

a_{1}+a_{3}+a_{5}=.9( a_{2}+a_{4}+a_{6})--------i

as Probability will not be greater than 1

therefore a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=1 ---------------------ii

probability of every even is equal (given)

a_{2}=a_{4}=a_{6}

put the value in equation i

a_{1}+a_{3}+a_{5}=.9( 3a_{2})

a_{1}+a_{3}+a_{5}=2.7(a_{2})----------------iii

put the value of equation (iii) in equation (ii)

then 2.7a_{2}+3a_{2}=1

5.7a_{2}=1

a_{2}=1/5.7=.1754338

Now If the probability that the face is even given that it is greater than 3 is 0.75

a_{4}+a_{6}/(a_{4}+a_{5}+a_{6}) =.75

2a_{2}/ (a_{4}+a_{5}+a_{6)}=.75 ( as a_{2}=a_{4}=a_{6} )

2a_{2}/ .75= a_{4}+a_{5}+a_{6}

2*(.1754338)/.75= a_{4}+a_{5}+a_{6}

.467823= a4+a5+a6 (answer)

Let a

_{1},a_{2},a_{3},a_{4},a_{5},a_{6}are the six facesProbability of odd= 90% Probability of even

a

_{1}+a_{3}+a_{5}=.9( a_{2}+a_{4}+a_{6})--------ias Probability will not be greater than 1

therefore a

_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=1 ---------------------iiprobability of every even is equal (given)

a

_{2}=a_{4}=a_{6}put the value in equation

ia

_{1}+a_{3}+a_{5}=.9( 3a_{2})a

_{1}+a_{3}+a_{5}=2.7(a_{2})----------------iiiput the value of equation (

iii) in equation(ii)then 2.7a

_{2}+3a_{2}=15.7a

_{2}=1a

_{2}=1/5.7=.1754338Now If the probability that the face is even given that it is greater than 3 is 0.75

a

_{4}+a_{6}/(a_{4}+a_{5}+a_{6}) =.752a

_{2}/ (a_{4}+a_{5}+a_{6)}=.75 ( as a_{2}=a_{4}=a_{6})2a

_{2}/ .75= a_{4}+a_{5}+a_{6}2*(.1754338)/.75= a

_{4}+a_{5}+a_{6}.467823= a4+a5+a6 (answer)