Option C
$P(\text{No two chocolates are identical}) = \frac{10\times9\times8\times7\times6}{{10}^5} = \frac{30240}{{10}^5} = 0.3024$
$P(\text{At least two chocolates are identical}) = 1 – P(\text{No two chocolates are identical})$
$\qquad \qquad = 1 – 0.3024 = 0.6976$
Alternatively,
Number of ways of selecting $5$ distinct chocolates, one each from the $5$ bags is same as selecting $5$ chocolates from $10$ distinct ones $ = {}^{10}C_5.$
If “distinct” requirement is not there, each of the $5$ chocolate has $10$ options $\implies 10^5.$
So, probability that no two chocolates are identical $ = \dfrac{{}^{10}C_5}{10^5} = 0.3024$
Probability that at least $2$ chocolcates are identical $ = 1 – 0.3024 = 0.6976$