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2 answers:
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Answer: 1. P = 1/64
2. P = 1/32
3. P = 1/8
Step-by-step explanation:
In genetics, an ofspring inherits one copy of an allele of each parent, which can be described in the tables below called Punnett Square:
For crossing Aa x Aa:
A a
A AA Aa
a Aa aa
For crossing Bb x Bb:
B b
B BB Bb
b Bb bb
For crossing Cc x Cc:
C c
C CC Cc
c Cc cc
We can separate them because they are assorted independently.
For offspring with <u>genotype</u> <u>AABBcc</u>, probability will be:
P(AA) = 1/4
P(BB) = 1/4
P(cc) = 1/4
As all three probabilities has to happen at the same time, it is a "E" rule:
P(AABBcc) =
P(AABBcc) = 1/64
Probability for the individual to be AABBcc is 1/64 or 1.56%.
<u>Genotype</u> <u>AaBBcc</u>:
P(Aa) = 2/4 = 1/2
P(BB) = 1/4
P(cc) = 1/4
P(AaBBcc) =
P(AaBBcc) = 1/32
Probability for the individual to be AaBBcc is 1/32 or 3.12%
<u>Genotype</u> <u>AaBbCc</u>:
P(Aa) = 1/2
P(Bb) = 1/2
P(Cc) = 1/2
P(AaBbCc) =
P(AaBbCc) = 1/8
Probability for the individual to be AaBbCc is 1/8 or 12.5%.
Answer:
We are given that Commercial aircraft used for flying in instrument conditions are required to have two independent radios instead of one.
The probability of failure of one radio = 0.0026
So, the probability of failure of both radio =
Now we are supposed to find the probability that a particular flight will be safe with at least one working radio
A) the probability that a particular flight will be safe with at least one working radio = 1- 0.00000676=0.99999324
B)The usual rounding rule of three significant digits not work here because usual rounding rule of three significant digits makes it 1 which means that both radios are working properly.
C)Yes,this probability is high enough to ensure flight safety because it is close to 1