Five less than 6 times a number plus seven
2 answers:
You might be interested in
Answer:
(a) The tree diagram is attached as an image
(b) P(NO ∩ D ∩ NA) = 0.27
(c) P(O ∩ C ∩ A) = 0.004
(d) P(A) = 0.438
(e) P(A|C) = 0.162
(f) P(C|A) = 0.374
Step-by-step explanation:
Let C denote cats and D denote dogs. Then,
P(C) = 40% = 0.4 and
P(D) = 60% = 0.6
Let O denote that an animal is older and NO denote that an animal is not older. Then for Cats,
P(O) = 20% = 0.2 and
P(NO) = 1 - 0.2 = 0.8
Similarly, for Dogs,
P(O) = 10% = 0.1 and
P(NO) = 1 - 0.1 = 0.9
Let A denote that an animal is adopted and NA denote that an animal is not adopted. For older cats,
P(A) = 5% = 0.05 and
P(NA) = 1 - 0.05 = 0.95
Similarly, for older dogs,
P(A) = 10% = 0.1 and
P(NA) = 1 - 0.1 = 0.9
For cats that are not older,
P(A) = P(NA) = 0.5
For dogs that are not older
P(A) = P(NA) = 0.5
(a) We can illustrate this information using a tree diagram as shown in the image attached. The first stage represents whether the animal is a cat or a dog. Then, they are classified as older or not older and then they are classified according to if they are adopted or not.
(b) We need to compute the probability that the selected animal is not older and a dog and is not adopted i.e. P(NO ∩ D ∩ NA). For this we will use the tree diagram and trace out the path D to NO to NA. We need this sample point hence, we will multiply all three probabilities and find out the answer.
P(NO ∩ D ∩ NA) = (0.6)*(0.9)*(0.5)
P(NO ∩ D ∩ NA) = 0.27
(c) Now, we need to compute P(O ∩ C ∩ A). So, using the tree diagram, we will trace out the route from C to O to A and multiply these three probabilities to get our answer.
P(O ∩ C ∩ A) = 0.4 * 0.2 * 0.05
P(O ∩ C ∩ A) = 0.004
(d) To compute the probability that a randomly chosen animal is adopted, we need to see the combinations in our tree diagram where we find adopted (A). The possible combinations can be: (C, O, A), (C, NO, A), (D, O, A) and (D, NO, A). So, the probability of adopted is:
P(A) = (C, O, A) + (C, NO, A) + (D, O, A) + (D, NO, A)
= 0.4*0.2*0.05 + 0.4*0.8*0.5 + 0.6*0.1*0.1 + 0.6*0.9*0.5
= 0.002 + 0.16 + 0.006 + 0.27
P(A) = 0.438
(e) Now, we are given that the chosen animal is a cat and we need to compute the probability that it is adopted. So, from the tree diagram, we can see that a cat who is oler can be adopted (C, O, A) and a cat who is not older can also be adopted (C, NO, A). These two sample points make up that probability that a cat is adopted. So,
P(A|C) = (C, O, A) + (C, NO, A)
= 0.4*0.2*0.05 + 0.4*0.8*0.5
= 0.002 + 0.16
P(A|C) = 0.162
(f) P(C|A) = P(C∩A)/P(A)
Now this can hold true for older as well as not older cats. We will consider both and add them.
P(C|A) = P(C∩A)/P(A) (older cats) + P(C∩A)/P(A) (not older cats)
= (0.4*0.2*0.05)/0.438 + (0.4*0.8*0.5)/0.438
= 0.004/0.438 + 0.16/0.438
= 0.009132 + 0.36529
P(C|A) = 0.374
