Question

Please help!!

A box contains 6 nuts, 8 bolts, and 4 screws. If 3 objects are selected in succession randomly, what is the probability of selecting a nut, then a bolt, then a screw, if replacement occurs each time?

A. 8/243 B. 16/243 C. 1/27 D. 1

Answer 1

Hint:

Assuming the selections are independent, you can say

P(Nut, bolt, screw) = P(Nut)*P(bolt)*P(screw)

Answer 2

let me know if this helps or not

Answer 3

so it would be P(6/17) * P(8/17) * P(4/17)?

Answer 4

lose the P's and turn all the 17's into 18

There are 6+8+4 = 18 items total (and not 17)

Answer 5

so you would have

(6/18) * (8/18) * (4/18)

as your next step

Answer 6

Okay and then you get 192/5832

Answer 7

now reduce as much as possible

Answer 8

3/729?

Answer 9

no

Answer 10

oh wait it would be 3/3 = 1

Answer 11

I would divide by 2 first

192/2 = 96

5832/2 = 2916

So 192/5832 = 96/2916

we can go further

Answer 12

96/2 = 48

2916/2 = 1458

So

96/2916 = 48/1458

Answer 13

Since 2 works a bunch of times, let's try 4

48/4 = 12

1458/4 = 364.5

so 4 doesn't work, so let's try 2 again

-------------------------------------------------------

Anyways, you do this trial and error a bunch of times

The best way to do it really is to find the GCF of 192 and 5832, which is 24, and divide each term by that

192/24 = 8
5832/24 = 243

Answer 14

So in the end,

192/5832

reduces to

8/243

Answer 15

Oh okay I get it now. Thank you

Answer 16

you're welcome