Because of a manufacturing error, 3 cans of regular soda were accidentally filled with diet soda and placed into a 24-pack. Suppose that two cans are randomly selected from the 24-pack. Determine the probability that at least one contain regular soda.

161/184 or 0.875

Step-by-step explanation:

Total number of cans = 24 cans

Total number of diet soda = 3 cans

Total number of regular soda = 21 cans

We are asked to find the probability that:that at least one contain regular soda if two cans are selected randomly

We have two ways for this happening

a) two of the cans are regular soda

b) one of the cans is regular , while one is diet

Hence,

Probability (that at least one contain regular soda) = Probability(that two of the cans are regular soda) + Probability ( one of the cans is regular , while one is diet)

Probability(that two of the cans are regular soda) = 21/24 × 20/23

= 35/46

Probability ( one of the cans is regular , while one is diet) = 21/24 × 3/23

= 21/184

Probability (that at least one contain regular soda) = 35/46 + 21/184

We find the Lowest common multiple of the denominators = 184

= 35/46 + 21/184

= (35 × 4) + (21 × 1)/184

= 140 + 21/184

= 161/184

= 0.875

Therefore, the probability that at least one can contains regular soda = 161/184 or 0.875

correct answer:

i^31 = -i