To determine the probabilities, we need to know the total number of pictures that show either Amy's cats, dogs, or both. To find this, we add the number of pictures showing only her cats (8), only her dogs (36), and both her cats and dogs (16): 8 + 36 + 16 = 60.
A. The probability of randomly selecting a picture that shows both her cats and dogs is 16/60 or approximately 26.7%, not 16%, so statement A is false.
B. The information given does not provide enough information to determine the probability of randomly selecting a picture that shows Amy, so statement B cannot be determined.
C. The information given does not provide enough information to compare the probabilities of randomly selecting a picture that shows Amy with her dogs and Amy with her cats, so statement C cannot be determined.
D. The probability of randomly selecting a picture that shows Amy’s dogs is 36/60 or 60%, and the probability of randomly selecting a picture that shows Amy’s cats is 8/60 or approximately 13.3%. Therefore, statement D is false.
None of the statements provided are true based on the given information.
Amy has 60 pictures of her pets in a digital album called “Amy’s Fur Babies” on her social media account. Of these pictures, 16 show both her dogs and cats, 8 show only her cats, and 36 show only her dogs.
Amy is in 1 of the pictures that show only her cats and 1 of the pictures that show
2 4only her dogs.
Based on this information, which statement is true?
Responses
A The probability of randomly selecting a picture that shows both her cats and dogs is 16%.The probability of randomly selecting a picture that shows both her cats and dogs is 16%.
B The probability of randomly selecting a picture that shows Amy is less than 20%.The probability of randomly selecting a picture that shows Amy is less than 20%.
C The probability of randomly selecting a picture that shows Amy with her dogs is greater than the probability of randomly selecting a picture that shows Amy with her cats.The probability of randomly selecting a picture that shows Amy with her dogs is greater than the probability of randomly selecting a picture that shows Amy with her cats.
D The probability of randomly selecting a picture that shows Amy’s dogs is 3 times the probability of randomly selecting a picture that shows Amy’s cats.The probability of randomly selecting a picture that shows Amy’s dogs is 3 times the probability of randomly selecting a picture that shows Amy’s cats.
3 answers
what would be the closest to true
Based on the given information, the closest statement to being true is:
B. The probability of randomly selecting a picture that shows Amy is less than 20%.
Since the number of pictures that show Amy is not explicitly given, we can assume that it is less than or equal to 2 (given that Amy appears in at most one picture for each category). Since there are 60 pictures in total, the probability of randomly selecting a picture that shows Amy is 2/60 or approximately 3.3%. Thus, the statement B is the closest to being true.
B. The probability of randomly selecting a picture that shows Amy is less than 20%.
Since the number of pictures that show Amy is not explicitly given, we can assume that it is less than or equal to 2 (given that Amy appears in at most one picture for each category). Since there are 60 pictures in total, the probability of randomly selecting a picture that shows Amy is 2/60 or approximately 3.3%. Thus, the statement B is the closest to being true.
1 answer