A two-digit locker combination is made up of nonzero digits and no digit is repeated in any combination.
Event A = the first digit is 1
Event B = the second digit is even
If a combination is picked at random with each possible locker combination being equally likely, what is P(A and B) expressed in simplest form?
Sorry the Answer is C
C.) 1/2
I found out the hard way :(
surely you can see that
P(A) = 1/9
P(B) = 4/10
so, what do you think?
Note that saying no digit is repeated is redundant, since A and B are mutually exclusive anyway.
Answer is A
To solve this problem, we need to find the probability of both Event A and Event B occurring.
Event A: The first digit is 1.
Since there are a total of 90 two-digit combinations without repetition (ranging from 10 to 98), there are 9 possible combinations where the first digit is 1 (10, 12, 13, ..., 19). Therefore, the probability of Event A occurring is 9/90 or 1/10.
Event B: The second digit is even.
Since there are a total of 90 two-digit combinations without repetition, half of them will have an even second digit. So, there are 45 possible combinations where the second digit is even. The probability of Event B occurring is 45/90 or 1/2.
To find the probability of both A and B occurring (A and B are independent events), we multiply the probabilities of each event since the occurrence of one event does not affect the occurrence of the other event.
P(A and B) = P(A) * P(B) = (1/10) * (1/2) = 1/20
Therefore, the probability of both Event A and Event B occurring is 1/20.