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

bryson has twelve cards. each card has a different number from 1 to 12 on it. he chooses a card at random. how likely is it for bryson to choose an odd number? as likely as not how likely is it for bryson to choose a number greater than 4? unlikely 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. how likely is it for bryson to choose an odd number? as likely as not how likely is it for bryson to choose a number greater than 4? unlikely 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 out of 12 cards. The probability of choosing an odd - numbered card is $P(\text{odd})=\frac{6}{12}=\frac{1}{2}$. Since the probability is $\frac{1}{2}$, it is as likely as not.

Step2: Count numbers greater than 4

The numbers greater than 4 from 1 to 12 are 5, 6, 7, 8, 9, 10, 11, 12. There are 8 such numbers out of 12 cards. The probability of choosing a number greater than 4 is $P(\text{>4})=\frac{8}{12}=\frac{2}{3}$. Since $\frac{2}{3}> \frac{1}{2}$, it is likely.

Answer:

How likely is it for Bryson to choose an odd number? as likely as not How likely is it for Bryson to choose a number greater than 4? likely