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A two-digit locker combination has two nonzero digits and no two digits are the same. Event A is defined as choosing an odd digit for the first number, and event B is defined as choosing an even digit for the second number. If a combination is picked at random, with each possible locker
combination being equally likely, what is P(B|A) expressed in simplest
form?

A) 1/9
B) 1/8
C)1/2
D) 5/9

Answer :

Answer:

Event A = 5/9

Event B = 4/8

we multiply these to get 5/9 x 4/8 = 20/72 = EVEN ODD

and 4/9 x 4/8  = 16/72  ODD EVEN

5/9 X 5/ 8 = 25/72 =  ODD ODD

4/9 X 3/8 = 12/72 = EVEN EVEN

Therefore; Random =

P) (20 + 16 + 25 + 12) / 4(72) = 72/ 4(72) =72/ 288 = 0.25 = 1/4

Step-by-step explanation:

Answer:

MY name's Jeff and The answer's 5/9

Step-by-step explanation:

The amount of odd numbers are 5 and out of the 9 you have a chance of getting one, so you would have a 5/9 chance

Tell me if I'm incorrect

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