Let, #M, and N# be the events that the chocolate selected
from the Box is a Milk one, and contains Nuts, resp.
Then, #M'# denotes the event that the chocolate selected is
not a milk chocolate.
To determine #P(M) :#
There are #(3x+4)+x+(x-2)+(2x+3)=(7x+5)#
chocolates in the box, out of which #1# chocolate can be selected in
#(7x+5)" ways"#.
No. of milk chocolates is #(3x+4)+x=(4x+4)#, so, #1# can be
chosen in #(4x+4)" ways"#.
#:. P(M)=(4x+4)/(7x+5)..................................(square_1)#.
Similarly, #P(N)=(2x-2)/(7x+5)...............................(square_2)#.
Given taht, #P(M)=3P(N) rArr (4x+4)/(7x+5)=3((2x-2)/(7x+5))#.
#rArr 4x+4=6x-6 rArr 10=2x, i.e., x=5#.
Part (a) :
#"No. of chocolates in the Box"=7x+5=7*5+5=40#.
Part (b) :
#"The Reqd. Prob.="(P(N))/(P(M'))#,
#=(P(NnnM'))/(P(M'))#,
#=(P(N)-P(NnnM))/(1-P(M))#,
#={(2x-2)/(7x+5)-x/(7x+5)}-:{1-(4x+4)/(7x+5)}#,
#=(x-2)/(3x+1)=(5-2)/(3*5+1)......[because, x=5]#.
#rArr "The Reqd. Prob.="1/16#.
Enjoy Maths., and, Spread the Joy!