Solution

Let 

$N_i$ be the event that $i^{th}$ bearing is not defective. Since, the choice is made from large lot, these events are independent of each other and probability of these events would be same.

Let $A$ be the event that lot is accepted.

                                                    A=N1​∩N2​⋯∩N7​

If none of the lot has defective bearing,

$P(N_i)=1, ~ \forall i \in \{1, \dots, 7\}$ and $P(A)=1$

If $\dfrac{1}{10}$ bearings are defective in the lot,

$P(N_i)=1 -0.1=0.9, ~ \forall i \in \{1, \dots, 7\}$ and $P(A)=(0.9)^7 \approx 0.48$

If $\dfrac{1}{2}$ bearings are defective in the lot,

$P(N_i)=1 -0.5=0.5, ~ \forall i \in \{1, \dots, 7\}$ and $P(A)=(0.5)^7 \approx 0.008$

$Answer$

$1;0.48;0.008$

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$$A= N_1 \cap N_2 \dots \cap N_7$$