Hardy Weinberg Principle for the USMLE Step 1 

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I think cuz it’s the higher probability

yea, that's what i was trying to figure out. I kept getting .006645

@@fatimaserrato6929 He used .988 instead of .997, so I guess he made a mistake

Because total probability in AR diseases is the probability of both carrier parents passing down the mutant allele (1/2 X 1/2 = 1/4) X probability of mother being carrier X probability of father being carrier . If the question says both parents are carriers , then each parent has a probability of “1” . That would be 1/4 X 1X 1= 1/4 (Like you said) However this question requires us to calculate the probability of each parent being a carrier . Thus: 1/4 X 1/150 X 2/3 1/4 = the probability of each parent passing down the mutant allele (1/2 X1/2) 1/150 = probability of father being a carrier 2/3 = probability of mother being a carrier If the question said mother was a carrier , then It would be : 1/4 X 1/150 X 1 1/4 = the probability of each parent passing down the mutant allele (1/2 X1/2) 1/150 = probability of father being a carrier 1 = mother is a carrier Thus as long as the question doesn’t mention either parent being a carrier (then probability would be 1 ) , then it requires us to calculate the probability .

Ohh i got it! Thanks!

In first question husband was carrier 2pq = 2(0.9967×1/300),he took 0.9967=approximately as 1 so 2(1×1/300) = 1/150