bryson has twelve cards. each card has a different number from 1 to 12 on it. he chooses a card at…

bryson has twelve cards. each card has a different number from 1 to 12 on it. he chooses a card at random.\nhow likely is it for bryson to choose an odd number?\nunlikely as likely as not likely certain

bryson has twelve cards. each card has a different number from 1 to 12 on it. he chooses a card at random.\nhow likely is it for bryson to choose an odd number?\nunlikely as likely as not likely certain

Answer

Explanation:

Step1: Count odd - numbered cards

The odd numbers from 1 to 12 are 1, 3, 5, 7, 9, 11. So there are 6 odd - numbered cards.

Step2: Calculate probability

The probability $P$ of choosing an odd - numbered card is the number of favorable outcomes (choosing an odd - numbered card) divided by the number of total outcomes (choosing any card). The total number of cards is 12. So $P=\frac{6}{12}=\frac{1}{2}$.

Answer:

as likely as not