Question

Can anyone answer this question on Probability?

Answer 1

i know this anwer .. but i may answer u it tommorow ..

Answer 2

Given that total number of bolts = 30.

The total number of nuts = 40.

Therefore the total number of bolts and nuts = 70.

Given that 2 items are drawn = 70c2.

Given that Half of the bolts are rusted = 30 - 15

                                                                = 15.


Given that Half of the nuts are rusted = 40 - 20

                                                               = 20.


Therefore the total number of rusted items = 15 + 20

                                                                         = 35.

Given that two items are drawn = 35c2.



Let A be the event of getting an item that is rusted from the drawn item.

P(A) = 35c2/70c2


Let B be the event of getting an item that is a bolt from the drawn item.

P(B) = 30c2/70c2.


The 15 bolts are in both A and B.

P(A intersection B) = 15c2/70c2


Therefore the probability that either both are rusted or both are bolts

P(AUB) = 35c2/70c2 + 30c2/70c2 - 15c2/70c2

            = 595/2415 + 435/2315 - 105/2415

            = (595 + 435 - 105)/2415

            = 925/2415

            = 185/483

           
Hope this helps!