contestada

I asked my students about their pets. When i asked who has a cat, 16 students raised their hand. When i asked who has a dog, 18 students raised their hand. I noticed that 6 students had raised their hand twice and 4 students did not raise their hand at all.

If I choose a dog owner randomly, what is the probability they have a cat also?

Respuesta :

semsee45

Answer:

[tex]\sf Probability=\dfrac{1}{3}[/tex]

Step-by-step explanation:

Let C be the number of students who have a cat.

Let D be the number of students who have a dog.

Let N be the number of students who do not have a dog or a cat.

Let C∩D be the number of students who have a cat and a dog.

From the given information:

  • C = 16
  • D = 18
  • N = 4
  • C∩D = 6

Now, we can use the principle of inclusion and exclusion to find the total number of students:

[tex]\begin{aligned}\textsf{Total Students} &= \sf C + D - C \cap D + N\\&=16 + 18 - 6 + 4\\& = 32\end{aligned}[/tex]

Therefore, the probabilities are:

[tex]\sf P(C) = \dfrac{16}{32}[/tex]

[tex]\sf P(D) = \dfrac{18}{32}[/tex]

[tex]\sf P(N )= \dfrac{4}{32}[/tex]

[tex]\sf P(C\cap D) = \dfrac{6}{32}[/tex]

Now, if we choose a dog owner randomly, the probability that they also have a cat is given by:

[tex]\sf P(C\;|\;D) = \dfrac{P(C \cap D)}{P(D)}[/tex]

Substitute in the values:

[tex]\sf P(C\;|\;D) = \dfrac{\frac{6}{32}}{\frac{18}{32}}[/tex]

[tex]\sf P(C\;|\;D) = \dfrac{6}{18}[/tex]

[tex]\sf P(C\;|\;D) = \dfrac{1}{3}[/tex]

Therefore, the probability that a randomly chosen dog owner also has a cat is 1/3.

Otras preguntas