Respuesta :
Answer: The probability that a gene is of type A is 0.29.
Step-by-step explanation:
Let A be the event of type A.
Let B be the event of type B.
Let D be the event of dominant.
So, P(B and D) = 0.22
P(D|B)=0.31
So, Using the conditional probability, we get that
[tex]P(B|D)=\dfrac{P(B\cap D)}{P(B)}\\\\0.31=\dfrac{0.22}{P(B)}\\\\P(B)=\dfrac{0.22}{0.31}=0.71[/tex]
So, the probability that a gene is of type A is given by
[tex]P(A)=1-P(B)=1-0.71=0.29[/tex]
Hence, the probability that a gene is of type A is 0.29.
Answer:
The probability that a gene is of type A is 0.29 .
Step-by-step explanation:
We are given that a gene can be either type A or type B, and it can be either dominant or recessive.
Let event A = gene having type A
event B = gene having type B
event D = gene is dominant
event R = gene is recessive
We are given that if the gene is type B, then there is a probability of 0.31 that it is dominant i.e. P(D/B) = 0.31
Also, there is a probability of 0.22 that a gene is type B and it is dominant i.e. P(B [tex]\bigcap[/tex] D) = 0.22
Now, the conditional probability of P(D/B) is given by;
⇒ P(D/B) = [tex]\frac{P(B \bigcap D) }{P(B)}[/tex]
⇒ 0.31 = [tex]\frac{0.22 }{P(B)}[/tex] ⇒ P(B) = 0.22 ÷ 0.31 = 0.71
So, now Probability that a gene is of type A, P(A) = 1 - P(B) {because only gene of type A or type B is possible}
= 1 - 0.71 = 0.29 .
Hence, the probability that a gene is of type A is 0.29 .
