Q77 / 160 · TS EAMCET 2022 (18 Jul Shift 2) · Mathematics

Let A,B,C be three pairwise independent events of a random experiment. If PB¯∪C¯=12,PA>0,PB=b and PC=c, then PB¯∩C¯A=

  • A
    1+b-c
  • B
    2+b-c
  • Correct answer
    C
    32-b-c
  • D
    2-b-c

Solution

Given,

A, B & C are independent so,

PA∩B=PAPB

PB∩B=PBPC

PC∩A=PCPA

And given PB¯∪C¯=12

Now,

PB¯∩C¯A=PB¯∩C¯∩APA

Figure

By diagram we can see,

PB¯∩C¯∩A=PA−PA∩B−PA∩C+PA∩B∩C

Also, PB¯∪C¯=12

⇒1−PB∩C=12

⇒PBPC=12

Rewriting,

PB¯∩C¯∩A=PA−PAPB−PAPC+PAPBPC

⇒PB¯∩C¯∩A=PA1−PB−PC+12

⇒PB¯∩C¯∩APA=1−PB−PC+12

⇒PB¯∩C¯∩APA=32−b−C

So, PB¯∩C¯A=32−b−C

Hence, the correct answer is option: 'C'.

Relevant Questions