Show that in a group of four people it is more likely than not that two of them were born on the same day of the week.

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I've started the working :

Possible combinations : AB AC AD BC BD CD = 6

P(2 people born same day of week) = 1 - P(No people born on same day of week)

=1−((7/7)⋅(6/7)⋅(5/7)⋅(4/7))=1−(120/343)=223/343≈0.65

I'm not sure if this is correct and would really appreciate any working or answers.

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As N. F. Taussig commented, you have too many $7$s and should have

$$= 1 - \left({\frac{7}{7} \cdot \frac{6}{7} \cdot \frac{5}{7} \cdot \frac{4}{7}}\right) = 1 - \frac{120}{343} = \frac{223}{343} \approx 0.65%%$$