A two-digit locker combination is made up of two non-zero digits. Digits in a combination are not repeated and range from 3 through 8.
Event A = choosing an odd number for the first digit
Event B = choosing an odd number for the second digit
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. 1/5
B. 3/14
C. 5/18
D. 2/5
odd numbers 3 5 7
odd for first = 1/3
then for second 1/2
both 1/6
the answer is maybe D 2/5ths :)
thank you all for the help :)
but maybe they mean ANY odd number, not a particular one
odd 3,5,7
even 4, 6, 8
first one 3/6 = 1/2
second one 2/5
both 0.2
Well, if we look at the possible odd digits that can be chosen for the first digit, we have 3, 5, and 7. And for the second digit, we also have 3, 5, and 7. Since there are 3 odd digits out of a total of 6 possible digits (3 through 8), the probability of choosing an odd number for the first digit (A) is 3/6 or 1/2.
Now, considering the second digit, after choosing an odd number for the first digit, we are left with 5 possible digits (3, 4, 5, 6, and 8), out of which 3 are odd. Therefore, the probability of choosing an odd number for the second digit (B) is 3/5.
To find the probability of both events A and B happening, we simply multiply their probabilities: (1/2) * (3/5) = 3/10.
Therefore, the answer is not listed among the options. It seems like someone set this up just to mess with you. Classic prank!
To find the probability of two independent events happening together, we multiply the individual probabilities of each event.
Event A is choosing an odd number for the first digit. The odd numbers between 3 and 8 are 3, 5, and 7. So, there are 3 options for the first digit.
Event B is choosing an odd number for the second digit. After choosing the first digit, there are 4 remaining digits (3, 5, 7, and 8), but we cannot repeat the first digit. So, there are 3 options for the second digit.
Since each possible locker combination is equally likely, there are a total of 6 possibilities (3 options for the first digit multiplied by 2 options for the second digit, since we have already chosen one digit).
Therefore, the probability of Event A and Event B happening together (P(A and B)) is (3/6) * (3/5) = 9/30 = 3/10.
However, the question asks for the probability to be expressed in simplest form. We can simplify 3/10 by dividing both numerator and denominator by their greatest common divisor, which is 1. So, the simplified form is 3/10.
Therefore, the correct answer is not provided in the options. The correct answer should be 3/10.