Female III-4 is pregnant via male III-5. The owner of this breeding pair wants to know the probabilities of several possible outcomes for their offspring (IV-3). If you need help with how to approach calculating these probabilities, use the Hint.

the following are required for IV-3 to have condition

Explanation:

-II-4 passes an X b chromosome to III-4 (probability = 1/2).

-If III-4 has the genotype X B X b (accounted for by the above probability), then she passes an X b chromosome to IV-3 (probability = 1/2).

-III-5 passes a Y chromosome to IV-3 (probability = 1/2).

All of these requirements are needed in sequence, so you apply the product rule here, too (1/2 x 1/2 x 1/2 = 1/8).

Once the individual probabilities are known, the sum and/or product rules can be used for various combinations (both conditions, either condition, etc.).

if the probability is 1/12, 75 people will receive a $25 gift card

step-by-step explanation: