Probability of presence of at least one color blind person in a group

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The frequency of red-green color vision defects among populations of Northern European origin is around 8% of males and 0.5% of females.

  1. Deeb S. The molecular basis of variation in human color vision: Variation in human color vision. Clinical Genetics. 2005;67(5):369-377. doi:10/cn5s4g

One souce I found for the probability of at least one color blind person in a group of three men is (https://jfly.uni-koeln.de/color/): $0.22 = 1-(1-0.92)^3$

Does that source do the correct calculation? I think it should be $0.22= 1 -(1-0.08)^3$

How could I calculate the chance of presence of at least one color blind person in a group of two women and two men?

Would $0.16 = 1 - (0.92^2 * 0.995^2)$ be correct?

Thank you for your help.

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Yes, they're both correct. The source is calculating the probability of the opposite event (which is easier). Note $P(Event) = 1 - P(Opposite \space event)$. There you have it