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Ilia_Sergeevich [38]
3 years ago
11

Question One:

Mathematics
1 answer:
Jobisdone [24]3 years ago
8 0

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

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Step-by-step explanation:

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  1                     6

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There are six different ways to roll a 7.  If you do the same for all possible numbers, you will see that there are 36 total options when rolling two dice.  The first die has six options, and the second die also has 6 options.  6x6 = 36.  (This is the fundamental counting principal you have have leared in statistics)

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