BITSAT Mathematics Question (2023) — Solution

Step-by-step solution

P(B)= 3 5 , P(A B)= 1 2 and P(A B)= 4 5 P(A B)=P(A B) P(B)= 1 2 3 5 = 3 10 P ( A B )= P ( A )+ P ( B )- P ( A B ) P ( A )= 4 5 - 3 10 = 1 2 . P ( A ^ )=1- P ( A )= 1 2 We know, P ( A B )+ P ( A ^ B )= P ( B ) [as A B and A^ B are mutually exclusive events] 3 10 + P ( A ^ B )= 3 5 P ( A ^ B )= 3 5 - 3 10 = 3 10 Now, P ( A ^ B )= P ( A ^ )+ P ( B )- P ( A ^ B ) = 1 2 + 3 5 - 3 10 = 5+6-3 10 = 4 5 P (( A B )^ )=1- P ( A B )=1- 4 5 = 1 5 P (( A B )^ )+ P ( A ^ B )= 1 5 + 4 5 =1