Probability that 2 of the 10 chargers will be defective =0.35
Number of ways of selecting 10 chargers from 20 chargers is 20C10
20C10 = 184756
Number of ways of selecting 10 chargers from 20 = 184756
Number of ways of selecting 2 defective chargers from 5 defective chargers = 5C2
5C2 = 10
Since 2 defective chargers have been chosen, there remains 8 to choose
Number of ways of selecting 8 good chargers from 15 remaining chargers = 15C8
Number of ways of selecting 8 good chargers from 15 remaining chargers = 6435
Probability that 2 of the 10 will be defective =
(10x6435)/184756
Probability that 2 of the 10 will be defective = 64350/184756
Probability that 2 of the 10 chargers will be defective =0.35
Learn more on probability here: brainly.com/question/24756209
I think it’s the first one
Answer:
Step-by-step explanation:
A delivery of 4 boxes will contain 2 more defective lightbulbs than a delivery of 2 boxes.
In the first column the decimal is .26 and the percent is 26%. In the second column the fraction is 31/50 and the percent is 62%. In the last column the fraction is 9/20 and the decimal is .45. Hope this helps! ;)
