A locker combination consists of two nonzero digits, and each combination consists of different digits. Event A is defined as choosing an odd number as the first digit, and event B is defined as choosing an odd number as the second digit.
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?
A. 5/18
B. 4/9
C. 1/2
D. 5/9
P(B|A)
(5/9) x (4/8) = 20/72
Reducing... we have... 5/18
To find the probability of event A and event B occurring, we need to first determine the total number of possible combinations meeting the given conditions.
Event A is defined as choosing an odd number as the first digit. Out of the nine possible digits (1-9), five are odd (1, 3, 5, 7, 9).
Event B is defined as choosing an odd number as the second digit. However, since the combination consists of different digits, we have eight options left to choose from.
Therefore, the total number of possible combinations meeting these conditions is 5 * 8 = 40.
Now we can calculate the probability of A and B occurring. Since each possible locker combination is equally likely, the probability of any specific combination being chosen is 1/40.
To find the probability of both A and B occurring, we divide the number of successful outcomes (1) by the total number of possible outcomes (40):
P(A and B) = 1/40
However, we need to express this probability in simplest form. Since the numerator and denominator share no common factors other than 1, the probability cannot be simplified further.
Therefore, the answer is:
A. 1/40 (simplified form)
so.... ti would be 5/18?
So the digits can each be from 1 to 9, without repetition, of which 5 are odd
Your event is that both are odd
P(odd and odd) = (5/9)(4/8) = ....