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A box contains different colored paper clips. The probability of drawing two red paper clips from the box without replacement is 1/7 , and the probability of drawing one red paper clip is 2/5 . What is the probability of drawing a second red paper clip, given that the first paper clip is red?

A. 1/6


B. 5/14


C.2/3


D. 2/35


plz explain how you got the answer!

User Lestat
by
8.0k points

2 Answers

1 vote

Answer: 5/14 which is choice B

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How I got this answer:

Define the following events

A = event of picking a red paper clip on the first selection

B = event of picking a red paper clip on the second drawing

Replacement is not made.

Now onto the probabilities for each

P(A) = 2/5 = 0.4 is given to us as this is simply the probability of picking red on the first try

P(A and B) = probability of both events A and B happeing simultaneously = 1/7

P(B|A) = probability event B occurs, given event A has occured

P(B|A) = probability of selecting red on second selection, given first selection is red (no replacement)

P(B|A) = P(A and B)/P(A)

P(B|A) = (1/7) / (2/5)

P(B|A) = (1/7) * (5/2)

P(B|A) = (1*5)/(7*2)

P(B|A) = 5/14

So if event A happens, then the chances of event B happening is 5/14

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A more concrete example:

If we had 15 paperclips, and 6 of them were red, then

P(A) = (# of red)/(# total) = 6/15 = 2/5

P(B|A) = (# of red left)/(# total left) = (6-1)/(15-1) = 5/14

P(A and B) = P(A)*P(B|A) = (2/5)*(5/14) = 10/70 = 1/7

User Luan Tran
by
8.0k points
3 votes

Answer: B

Explanation:

Draw 1 (red) and Draw 2 (also red) = Both red


(2)/(5) * x =
(1)/(7)

Solve the equation to find the probability:


(2)/(5)x = (1)/(7)


((5)/(2))(2)/(5)x = ((5)/(2))(1)/(7)


x = (5)/(14)


User Bwegs
by
8.0k points