A two-digit locker combination has two nonzero digits and no two digits are the same. Event A is defined as choosing an odd digi

Question

3 years ago

10

A two-digit locker combination has two nonzero digits and no two digits are the same. Event A is defined as choosing an odd digi

t 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

Mathematics

2 answers:

Roman55 [17] · Answer 1 · 2022-10-30T03:40:01+03:00

Roman55 [17]3 years ago

8 0

Answer:

need the points sorry dude

Step-by-step explanation:

Nezavi [6.7K] · Answer 2 · 2022-10-30T06:03:02+03:00

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: