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Bb = brown hair phenotype
bb = black hair phenotype
we see that B = brown hair is the dominant gene while b = black hair is the recessive gene
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ll = short hair, this is a lowercase L
Ll = long hair
In this case, l = short hair is the recessive gene while L = long hair is the dominant gene.
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Check out the diagram below to see the dihybrid crosses for each of the hair color and hair length genotypes. From there, I show all of the genotypes possible for any offspring. Those genotypes are
- LlBb = long brown hair (comes up 4 times)
- Llbb = long black hair (comes up 4 times)
- llBb = short brown hair (comes up 4 times)
- llbb = short black hair (comes up 4 times)
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If we are talking about 1 offspring, then it is not possible for him/her to have both "long black hair AND short brown hair" as the instructions state. So let's consider two offspring instead.
The probability of having long black hair is 4/16 = 1/4 as there are 4 cases of Llbb out of 16 cases total. Similarly, the probability of getting llBb is 4/16 = 1/4 as well, for pretty much the same reasons.
The probability of having both events occur is (1/4)*(1/4) = 1/16
The multiplication of probabilities is valid because the two events are independent.
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<h3>Final Answer: A) 1/16</h3>
Student B is wrong , student A is correct in simplifying the expression .
Expression B must be simplified to cot θ .
Formulas to be used are :
1 - θ =
θ
1 - θ =
θ .
Since , location of error is step 3 , we will start from there
[ ( 1 - θ ) / sin θ ] / cos θ = [ ( 1 -
θ ) / sin θ ] * [ 1 / cos θ ]
= ( 1 - θ ) / ( sin θ * cos θ )
= θ / ( sin θ * cos θ )
= cot θ .
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Since , the question is incomplete we assume it to be :
