A two-digit locker combination has two non-zero digits and no digit is repeated in any combination
Event A = the first digit is less than 3
Event B = the second digit is less than 3
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. 2/9
D. 1/4
1/8
1/8 so the other dude is right sorry for being late 🙏
1/8 have a nice day
To find the probability of Event B given Event A, we need to find the probability of Event B occurring, assuming that Event A has already occurred.
Event A states that the first digit is less than 3, which means the possible first digits are 1 and 2. Since we know that no digit can be repeated, there are only two possible combinations for the first and second digits: (1, 2) and (2, 1).
Out of these two possible combinations, only one of them satisfies Event B, which states that the second digit is less than 3. The combination (1, 2) satisfies this condition, while (2, 1) does not.
Therefore, the probability of B given A (P(B|A)) is 1 out of 2, or 1/2.
However, we need to express the answer in simplest form. To do this, we can divide the numerator and denominator of 1/2 by their greatest common divisor, which is 1.
Dividing both numerator and denominator by 1, we get:
P(B|A) = 1/2
Therefore, the correct answer is not among the options provided.