A two-digit locker combination has two nonzero digits and no digit is repeated in the combination.

Event A=the first digit is less than 7.
Event B=the second digit is less than 7.

If a combination is picked at random with each possible locker combination being equally likely, what is P(B|A) expressed in simplest form?

P(B|A) = 0.9

Step-by-step explanation:

Given -

Event A=the first digit is less than 7 =   { 1,2,3,4,5,6 }

we cant choose 0 for first digit then the total no of outcome for first digit is 9

then P(A) =  

Event B=the second digit is less than 7 = {0, 1,2,3,4,5,6 }

then P(B)  =

=  { 1,2,3,4,5,6 }

P(B|A) =

            =

             = = .9

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

125 yards

step-by-step explanation:

i take 75 plus 50.