IIT JEE , AIEEE Forum - Mathematics , Algebra

danny_007

for three independent events A , & C , probability of A to occur is a , probability of none of them to occur is b and probability of atleast 1 of them not to occur is c .If p denotes the probability that C occurs but neither A nor B occurs , prove that p satisfies

ap² + [ab - ( 1 - a)(a + c - 1)]p + b(1 - a)( 1 - c) = 0

also show

c > (1 - a)² + ab / 1 - a

chimanshu_007

@ means intersection

p(A) = a

p(barA @ barB @ barC) = b

p (barA) p(barB) p(barC) = b

{1 - p(A)}{1 - p(B)}{1 - p(C)} = b

p bar(A @ B @ C) = c

1 - p(A @ B @ C) = c

1 - p(A)p(B)p(C) = c

p(barA) *p(barB)p(C)=p

{1 - p(A)}{1 - p(B)} p(C) = p

let p(A) = x , p(B) =y p(C) = z

so all the equations become

x = a

(1 - x)(1 - y)(1 - z) = b

1 - xyz = c

z(1 - x)(1 - y) = p

(1 - x)(1 - y)(1 - z)/z (1 - x)(1 - y) = b/p

1 - z / z = b/p

1 / z = (b + p) / p

z = p / (b+p)

z (1-x)(1-y) = p

1 - y = p/z(1-x)

y = 1 - p(b+p)/p(1-a)..........as x = a

y = 1 - a - b - p / (1 - a)

put the values of x , y , z in 1 - xyz = c

u will get the relation (1st one)

for the second part ,

since roots of the =n is real and atleast 1 must lie b/w 0 and 1

so

ab - (1 - a)(a + c - 1) < 0

ab + (1 - a)^2 - c ( 1-a) < 0

c > ab + (1-a)^2 / (1-a)

hence proved

danny_007

thanx

Ask & Discuss Questions with Community & Experts