Abox 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!

5/14 which is choice B

How I got this

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

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

B

Step-by-step explanation:

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

                       *                    x                    =        

Solve the equation to find the probability:

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !