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
Answer:

Step-by-step explanation:
Given

Now we know that system has infinite solution for x

in above equation.

∴
Answer: -6 is the answer
Step-by-step explanation:
Answer:
Binomial
There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.
Step-by-step explanation:
For each copy of the document, there are only two possible outcomes. Either it is defective, or it is not. This means that we can solve this problem using the binomial probability distribution.
Binomial probability distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
In this problem
Of the 20 copies, 2 are defective, so
.
What is the probability that you will encounter neither of the defective copies among the 10 you examine?
This is P(X = 0) when
.


There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.