A bag contains `2` red balls and `4` black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is red?

Explanation

(i) Put n=2 in `vec(OP)_(n-1)+ vec(OP) _(n+1)= (3)/(2) vec(OP)_n`
`" "vec(OP)_3= (3)/(2) vec(OP)_2- vec(OP)_1" "`(i)
`" "vec(OP)_1= ahati+ (1)`
`{:(vec(OP)_1= ahati+ (1)/(a) hatj),(vec(OP)_2= bhati+ (1)/(b)hatj):}}ab ne 0`
`therefore " "vec(OP)_3 = (3)/(2) (bhati+ (1)/(b) hatj)- (ahati+ (1)/(a) hatj)`
`" "= ((3b)/(2)-a) hati+ ((3)/(2b) - (1)/(a) )hatj`
If `P_3` lies on xy=1, we have
`" "((3b)/(2)-a)((3)/(2b)- (1)/(a))= 1`
or `" "(3b-2a)(3a-2b) = 4ab`
or `" "9ab-6b^(2) - 6a^(2) + 4ab = 4ab`
or `" "2a^(2) - 3ab+ 2b^(2)=0`
which is not possible as Discriminant `lt 0 (a= 0 and b=0` not possible)
(ii) `vec(OP) = cos alpha hati+ sinbeta hatj`
`" "vec(OP)_2 = cos beta hati+ sinbetahatj`
`therefore " "vec(OP)_3= (3)/(2) (cosbeta hati+ sinbeta hatj) - (cosalpha hati+ sin alpha hatj)`
`" "= ((3)/(2) cos beta - cos alpha)hati + ((3)/(2) sin beta-sinalpha) hatj`
Since `P_3` lies on `x^(2) + y^(2)=1`, we have
`((3)/(2) cos beta - cosalpha)^(2) + ((3)/(2) sinbeta-sinalpha)^(2)=1`
or `" "(9)/(4) + 1-3(cosbeta cosalpha + sin beta sinalpha )=1`
or `" "(9)/(4) -3 cos (beta-alpha)=0 rArr cos (beta -alpha)=(3)/(4)" "` (ii)
Put n=3 in the given relation.
`vec(OP)_2 + vec(OP)_4 = (3)/(2) vec(OP)_3`
or `" " vec(OP)_4 = (3)/(2) vec(OP)_3-vec(OP)_2`
`" "=(3)/(2) ((3)/(2) vec(OP)_2-vec(OP)_1)-vec(OP)_2`
`" "=(5)/(4) vec(OP)_2- (3)/(2) vec(OP)_1`
`" "= (5)/(4) (cosbeta hati+sin betahatj) - (3)/(2) (cosalpha hati+ sin alpha hatj)`,
Which lies on `x^(2)+y^(2)=1`