Help me please!!!!! A manufacturer produces soda cans and a quality control worker randomly selects two cans from the assembly line for testing. Past statistics show that 10% of the cans are defective. What is the probability that the two selected cans are defective if the quality control worker selects the two cans from a batch of 60 cans?

A. P(Both defective) = start fraction six over 25 end fraction
B. P(Both defective) = start fraction one over 118 end fraction
C. P(Both defective) = start fraction three over 250 end fraction
D. P(Both defective) = start fraction nine over 625 end fraction

Answers

Answer 1

Answer:

The probability that the two selected cans are defective, when the quality control worker selects the two cans from a batch of 60 cans is 1/118.

What is probability?

Probability of an event is the ratio of number of favorable outcome to the total number of outcome of that event.

A manufacturer produces soda cans and a quality control worker randomly selects two cans from the assembly line for testing.

Past statistics show that 10% of the cans are defective. Thus the probability of defective can to be selected is,

$P=(1)/(10)$

There is total 60 cans in which 10 percent are defective. Hence the total defective can in one batch are 6.

In the second attampt the number of total can in batch will be 59 and number of defective can will be 5.

The probability that the two selected cans are defective, when the quality control worker selects the two cans from a batch of 60 cans by chain rule is,

$P=(6)/(60)*(5)/(59)\nP=(1)/(118)$

Thus, the probability that the two selected cans are defective, when the quality control worker selects the two cans from a batch of 60 cans is 1/118.

Learn more about the probability here;

brainly.com/question/24756209

Answer 2

Answer: The probability that the two selected cans are defective if the quality control worker selects the two cans from a batch of 60 cans is P(Both defective) = start fraction nine over 625 end fraction. The answer is letter D

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