A two-digit locker combination is made up of non-zero digits and no digit is repeated in any combination.
Event A = the first digit is 4
Event B = the second digit is odd
If a combination is chosen at random with each possible locker combination being equally likely, what is P(A and B) expressed in simplest form?
A.
B.
C.
D.
To find the probability of Event A (the first digit is 4) and Event B (the second digit is odd), we need to consider the number of possible combinations that satisfy both events out of the total number of possible combinations.
Event A: The first digit is 4. Since only non-zero digits can be used, there are 9 choices for the first digit (1-9).
Event B: The second digit is odd. Since no digit can be repeated, there are 5 choices for the second digit (1, 3, 5, 7, 9).
To find the probability of both events occurring, we multiply the probabilities of each event happening:
P(A and B) = P(A) * P(B)
P(A) = 1/9 (probability of a 4 in the first digit)
P(B) = 5/9 (probability of an odd digit in the second digit)
P(A and B) = (1/9) * (5/9) = 5/81
Therefore, the correct answer is option C: P(A and B) = 5/81.
P(A) = 1/10
P(B) = P(A)*4/9 + (1-P(A))*5/9
so, what do you think?