Bag A contains 2 unbiased and 3 biased coins whereas Bag B contains 3 unbiased and 2 biased coins. A bag is selected at random and 2 coins are taken out simultaneously. The probability, that both coins are unbiased is:

Bag A contains 2 unbiased and 3 biased coins (5 coins total).

Bag B contains 3 unbiased and 2 biased coins (5 coins total).

A bag is selected at random, and 2 coins are taken out simultaneously.

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Since either bag can be selected:

$P(\text{both unbiased}) = P(\text{Bag A}) \times P(\text{both unbiased from A}) + P(\text{Bag B}) \times P(\text{both unbiased from B})$

The probability of selecting each bag is:

$P(\text{Bag A}) = \frac{1}{2}$

$P(\text{Bag B}) = \frac{1}{2}$

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For Bag A with 2 unbiased coins:

Number of ways to select 2 unbiased coins from 2: $\binom{2}{2} = 1$

Total ways to select 2 coins from 5: $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$

$P(\text{both unbiased | Bag A}) = \frac{1}{10}$

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For Bag B with 3 unbiased coins:

Number of ways to select 2 unbiased coins from 3: $\binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3$

Total ways to select 2 coins from 5: $\binom{5}{2} = 10$

$P(\text{both unbiased | Bag B}) = \frac{3}{10}$

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$P(\text{both unbiased}) = \frac{1}{2} \times \frac{1}{10} + \frac{1}{2} \times \frac{3}{10}$

$= \frac{1}{20} + \frac{3}{20}$

$= \frac{4}{20}$

$= \frac{1}{5}$

Therefore, the probability that both coins are unbiased is $\frac{1}{5}$.