Three students X , Y and Z appear for an examination. The probability of X passing the examination is 1 5 , the probability of Y passing the examination is 1 4 and the probability of Z failing the examination is 2 3 . The probability that atleast two of them pass the exam is

Step-by-step solution

P( pass )+P( fail )=1 aligned & P (x passing )+ P (x -fail ) =1 & P (x fail )=1- 1 5 = 4 5 =P (x^ ) & Similarly P (y fail )= P (y^ )=1- 1 4 = 3 4 & P (z pass )= P (z)=1- 2 3 = 1 3 aligned Probability of atleast 2 passing =( Any 2 pass and 1 fail ) + (All 3 pass) aligned & = P (x) P (y) P (z^ )+ P (x) P (y^ ) P (z)+ P (x^ ) P (y) & P ( z )+ P (x) P (y) P ( z ) aligned aligned & = ( 1 5 1 4 2 3 )+ ( 1 5 3 4 1 3 )+ ( 4 5 1 4 1 3 )+ ( 1 5 1 4 1 3 ) & = 2 60 + 3 60 + 4 60 + 1 60 = 10 60 = 1 6 aligned