MHB Probability that a randomly selected chicken has the genetic modification

[mathmari]

Hey! :giggle:

In the case of the reproduction of the chicken of the breed Krueper, in some cases it can be that they are extremely short-legged, which has as a result that they already die in the egg. A group of researchers now considers that they have found two main genetic modifications, which have a short-leg effect. These two modifications never occur at the same time. The genetic modification 1 is observed twice as often as genetic modification 2, but in 70% of the cases none of the two. In total, the short-leggedness occurs in 21% of all cases. If the genetic modification 1 occurs, the chicked are in 80% of the cases short-legged, and if the genetic modification 2 occurs, it is 40%.

(a) State all probabilities contained in the text. Introduce suitable events for this purpose.
(b) Determine the probability that a randomly selected chicken has the genetic modification 1. Determine also the probability that the genetic modification 2 is present in a randomly selected chicken.
(c) A chicken chosen at random is short-legged. What is the probability with which genetic modification 1 is present?
(d) What is the probability that a chicken is short-legged even though neither the first nor the second genetic modification is present?
I have done the following :For (a) :

We consider the following events :

• $K$: A randomly selected chicken is short-legged
• $V_1$ : The genetic modification 1 exists.
• $V_2$ : The genetic modification 1 exists.

From the above text we get the below probabilities :

The genetic modification 1 is observed twice as often as genetic modification 22 : $P(V_1)=2\cdot P(V_2)$
In 70% of the cases none of the two genetic modifications occur : $P((V_1\cup V_2)^c)=0.70$
In total, the short-leggedness occurs in 21% of all cases : $P(K)=0.21$
If the genetic modification 1 occurs, the chicked are in 80% of the cases short-legged : $P(K\mid V_1)=0.80$
If the genetic modification 1 occurs, the chicked are in 40% of the cases short-legged: $P(K\mid V_2)=0.40$Is everything correct and complete so far? :unsure:

For (b) :

We have the below :
$P((V_1\cup V_2)^c)=0.70 \Rightarrow P(V_1^c\cap V_2^c)=0.70$
$P(K\mid V_1)=0.80 \Rightarrow \frac{P(K\cap V_1)}{P(V_1)}=0.80$
$P(K\mid V_2)=0.40 \Rightarrow \frac{P(K\cap V_2)}{P(V_2)}=0.40$

The probability that a randomly selected chicken has the genetic modification 1 ia equal to $P(V_1)$, right?
Respectively, the probability that the genetic modification 2 is present in a randomly selected chicken is $P(V_2)$, right?

:unsure:

 

[I like Serena]

For (a) :
The genetic modification 1 is observed twice as often as genetic modification 22 : $P(V_1)=2\cdot P(V_2)$
In 70% of the cases none of the two genetic modifications occur : $P((V_1\cup V_2)^c)=0.70$
In total, the short-leggedness occurs in 21% of all cases : $P(K)=0.21$
If the genetic modification 1 occurs, the chicked are in 80% of the cases short-legged : $P(K\mid V_1)=0.80$
If the genetic modification 1 occurs, the chicked are in 40% of the cases short-legged: $P(K\mid V_2)=0.40$

Hey mathmari!

I'm missing: "These two modifications never occur at the same time."

For (b) :
$P((V_1\cup V_2)^c)=0.70 \Rightarrow P(V_1^c\cap V_2^c)=0.70$
$P(K\mid V_1)=0.80 \Rightarrow \frac{P(K\cap V_1)}{P(V_1)}=0.80$
$P(K\mid V_2)=0.40 \Rightarrow \frac{P(K\cap V_2)}{P(V_2)}=0.40$
The probability that a randomly selected chicken has the genetic modification 1 ia equal to $P(V_1)$, right?
Respectively, the probability that the genetic modification 2 is present in a randomly selected chicken is $P(V_2)$, right?

Yes... but it looks as if we need one more piece of information...

 

[mathmari]

I'm missing: "These two modifications never occur at the same time."

Ah yes! From that we get $P(V_1\cap V_2)=0$, or not? :unsure:

Yes... but it looks as if we need one more piece of information...

How do we get the information that we need? :unsure:

 

[I like Serena]

Ah yes! From that we get $P(V_1\cap V_2)=0$, or not?

How do we get the information that we need?

Indeed. And since $V_1$ and $V_2$ are mutually exclusive, we can apply the sum rule on the probability of their union
That should give us the information we need.

 

[mathmari]

Indeed. And since $V_1$ and $V_2$ are mutually exclusive, we can apply the sum rule on the probability of their union
That should give us the information we need.

We have that \begin{align*} P((V_1\cup V_2)^c)=0.70&\Rightarrow 1-P(V_1\cup V_2)=0.70\\ & \Rightarrow P(V_1\cup V_2)=0.30\\ & \Rightarrow P(V_1)+P(V_2)=0.30\\ & \Rightarrow 3P(V_2)=0.30\\ & \Rightarrow P(V_2)=0.10\end{align*} and so $P(V_1)=0.20$.

At (c) we want to calculate the probability $P(V_1\mid K) $. We calculate that using the formula $\frac{P(K\mid V_1)P(V_1)}{P(K)}$ where everything is now known.

Is everything correct so far? :unsure: At (d) do we want to calculate the probability $P(K\mid (V_1\cup V_2)^c)$?

:unsure:

 

[I like Serena]

All correct. (Nod)

 

[mathmari]

All correct. (Nod)

So at (d) we have $$P(K\mid (V_1\cup V_2)^c)=\frac{P(K\cap (V_1\cup V_2)^c)}{P((V_1\cup V_2)^c)}= \frac{P(K\cap (V_1\cup V_2)^c)}{0.70}$$But how can we calculate the numerator? It is equal to $P(K\cap (V_1\cup V_2)^c)=P(K\cap V_1^c\cap V_2^c)$. This means that teh chicken is short-legged but neither genetic modification 1 nor genetic modification occur, right? Is this equal to $0,70$ ? :unsure:

 

[I like Serena]

So at (d) we have $$P(K\mid (V_1\cup V_2)^c)=\frac{P(K\cap (V_1\cup V_2)^c)}{P((V_1\cup V_2)^c)}= \frac{P(K\cap (V_1\cup V_2)^c)}{0.70}$$But how can we calculate the numerator? It is equal to $P(K\cap (V_1\cup V_2)^c)=P(K\cap V_1^c\cap V_2^c)$. This means that teh chicken is short-legged but neither genetic modification 1 nor genetic modification occur, right? Is this equal to $0,70$ ?

We know that $P(K)=21\%$ and we also have $K=(K\cap V_1)\cup(K\cap V_2)\cup(K\cap (V_1\cup V_2)^c)$ with parts that are mutually exclusive.

 

[mathmari]

We know that $P(K)=21\%$ and we also have $K=(K\cap V_1)\cup(K\cap V_2)\cup(K\cap (V_1\cup V_2)^c)$ with parts that are mutually exclusive.

Ah ok! And so we get \begin{align*}P(K)=P(K\cap V_1)+P(K\cap V_2)+P(K\cap (V_1\cup V_2)^c) & \Rightarrow 0.21=0.80\cdot 0.20+0.40\cdot 0.10+P(K\cap (V_1\cup V_2)^c) \\ & \Rightarrow 0.21=0.16+0.04+P(K\cap (V_1\cup V_2)^c) \\ & \Rightarrow 0.21=0.20+P(K\cap (V_1\cup V_2)^c) \\ & \Rightarrow P(K\cap (V_1\cup V_2)^c)=0.01 \end{align*} Therefore we get $$
P(K\mid (V_1\cup V_2)^c)=\frac{P(K\cap (V_1\cup V_2)^c)}{P((V_1\cup V_2)^c)}= \frac{0.01}{0.70}\approx 0.014$$ Is everything correct? :unsure:

 

[I like Serena]

Ah ok! And so we get \begin{align*}P(K)=P(K\cap V_1)+P(K\cap V_2)+P(K\cap (V_1\cup V_2)^c) & \Rightarrow 0.21=0.80\cdot 0.20+0.40\cdot 0.10+P(K\cap (V_1\cup V_2)^c) \\ & \Rightarrow 0.21=0.16+0.04+P(K\cap (V_1\cup V_2)^c) \\ & \Rightarrow 0.21=0.20+P(K\cap (V_1\cup V_2)^c) \\ & \Rightarrow P(K\cap (V_1\cup V_2)^c)=0.01 \end{align*} Therefore we get $$
P(K\mid (V_1\cup V_2)^c)=\frac{P(K\cap (V_1\cup V_2)^c)}{P((V_1\cup V_2)^c)}= \frac{0.01}{0.70}\approx 0.014$$ Is everything correct?

You may want to add the intermediate step that $P(K\cap V_1)+\ldots=P(K\mid V_1)\cdot P(V_1)+\ldots=0.80\cdot 0.20+\ldots$.
Other than that, it looks correct to me. (Sun)

 

[mathmari]

You may want to add the intermediate step that $P(K\cap V_1)+\ldots=P(K\mid V_1)\cdot P(V_1)+\ldots=0.80\cdot 0.20+\ldots$.
Other than that, it looks correct to me. (Sun)

Great! Thanks a lot for your help!