A factory has three machines that manufacture widgets. The percentages of a tota

Question

A factory has three machines that manufacture widgets. The percentages of a total day's production manufactured by the machines are 10%, 35%, and 55%, respectively. Further, it is known that 5%, 3%, and 1% of the outputs of the respective three machines are defective. What is the probability that a randomly selected widget at the end of the day's production runs will be defective?

Question 3 options:

1.5%

2.0%

2.1%

2.5%

If P(A) = 0.46, P(B) = 0.4, and events A and B are independent, find the probability that neither A nor B happens.

Question 7 options:

0.184

0.274

0.216

0.324

Question 8 (1 point)

The probability that a coin lands on heads is 0.6. What is the probability that the coin does not land on heads when being independently tossed for 4 times?

Question 8 options:

0.2560

0.1296

0.01296

0.0256

Question 9 (1 point)

If P(C) = 0.2, P(B|C) = 0.25, and P(A|B and C) = 0.50, find the probability that events A, B, and C happen at the same time.

Question 9 options:

0.013

0.018

0.025

0

1.5%

2.0%

2.1%

2.5%

If P(A) = 0.46, P(B) = 0.4, and events A and B are independent, find the probability that neither A nor B happens.

Question 7 options:

0.184

0.274

0.216

0.324

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Question 8 (1 point)

The probability that a coin lands on heads is 0.6. What is the probability that the coin does not land on heads when being independently tossed for 4 times?

Question 8 options:

0.2560

0.1296

0.01296

0.0256

Save

Question 9 (1 point)

If P(C) = 0.2, P(B|C) = 0.25, and P(A|B and C) = 0.50, find the probability that events A, B, and C happen at the same time.

Question 9 options:

0.013

0.018

0.025

0

Explanation / Answer

Ans:

1)

P(defective)=P(defective/A)*P(A)+P(defective/B)*P(B)+P(defective/C)*P(C)

=0.05*0.1+0.03*0.35+0.01*0.55

=0.021 or 2.1%

Correct option is 2.1%

2)For independent events,P(A and B)=P(A)*P(B)

Now,

P(A or B)=P(A)+P(B)-P( A and B)

=0.46+0.4-0.46*0.4

=0.676

P(neither A nor B)=1-0.676=0.324

3)

P(not heads)=P(all tails)=(1-0.6)^4=0.4^4=0.0256

4)

P(B/C)=0.25

P(B and C)/P(C)=0.25

P(B and C)=0.25*0.2=0.05

Now,

P(A / B and C)=0.50

P(A and B and C)/P(B and C)=0.50

P(A and B and C)=0.50*0.05=0.025

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