Let S represents the sample space containing all possible ways of selecting 2 items (out 0f 70 nuts and bolts)

Let R represents the event of drawing 2 rusted item.

And B be the event of drawing 2 bolts

According to question-

n(S) = total ways of selecting 2 items out of 70 = ⁷⁰C₂

∵ half of bolts (30/2 = 15) and half of nuts(40/2 = 20) are rusted

∴ n(R) = no of ways in which 2 rusted items can be drawn.

⇒ n(R) = ³⁵C₂

And n(B) = no of ways in which 2 bolts can be drawn.

∴ n(B) = ³⁰C₂

Also, n(B ∩ R) = no of ways of selecting 2 items such that they are rusted bolts

∴ n(B ∩ R) = ¹⁵C₂

Hence, we have –

P(R) =

P(B) =

And P(R ∩ B) =

As we have to find probability of the events such that drawn items are rusted or bolts i.e. P(R ∪ B)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(R ∪ B) = P(R) + P(B) – P(R ∩ B)

⇒ P(R ∪ B) =