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Two parents each have blood type AB, and their first child is blood type AB. What is the probability that their second child will have blood type AB? ( Express the probability as a percent. Enter the number only without the percent sign. For example, enter 100% as 100 and enter 12.5% as 12.5 )

Respuesta :

Answer: 100% to 12.8

Step-by-step explanation:

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Using the principle of probability and the punnet square approach, the probability that the second child will have a blood type AB is 50%.

Drawing a Punnet square :

______ A ____ B

A ___ AA ____AB

|

B ___ AB ____BB

The total possible blood types = {AA, BB, AB, AB}

Probability = required outcome / Total possible outcomes

P(A) = 1/4 = 0.25 = 25%

P(B) = 1/4 = 0.25 = 25%

P(AB) = 2/4 = 0.50 = 50%

The probability of the blood group of each offspring is independent and does not rest on the bloodbgrouo or type of the children before or after.

Hence, the probability that the second child has a bllof tyo AB is 50%

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