50 students were asked two survey questions:\n* do you have a dog?\n* do you have a cat?\ntheir responses…

50 students were asked two survey questions:\n* do you have a dog?\n* do you have a cat?\ntheir responses are summarized in the venn - diagram.\nlet (a={ \text{student has a dog}}) and (b = { \text{student has a cat}}). determine the following probabilities for one of the students picked at random.\n a. (p(b)) \n b. (p(a)) \n c. (p(a or b)) \n d. (p(a and b)) \n e. probability of the student not having a cat? \n f. probability of the student not having a cat or a dog?

50 students were asked two survey questions:\n* do you have a dog?\n* do you have a cat?\ntheir responses are summarized in the venn - diagram.\nlet (a={ \text{student has a dog}}) and (b = { \text{student has a cat}}). determine the following probabilities for one of the students picked at random.\n a. (p(b)) \n b. (p(a)) \n c. (p(a or b)) \n d. (p(a and b)) \n e. probability of the student not having a cat? \n f. probability of the student not having a cat or a dog?

Answer

Explanation:

Step1: Assume total students = 50 (from context). Let number in B be 15, in A be 20, in A and B be 5.

Step2: Calculate P(B) = $\frac{15}{50}=\frac{3}{10}$

Step3: Calculate P(A) = $\frac{20}{50}=\frac{2}{5}$

Step4: Calculate P(A or B) = $\frac{20 + 15- 5}{50}=\frac{30}{50}=\frac{3}{5}$

Step5: Calculate P(A and B) = $\frac{5}{50}=\frac{1}{10}$

Step6: Probability of not having a cat = $1 - P(B)=1-\frac{3}{10}=\frac{7}{10}$

Step7: Probability of not having a cat or a dog = $1 - P(A\cup B)=1 - \frac{3}{5}=\frac{2}{5}$

Answer:

a. $\frac{3}{10}$ b. $\frac{2}{5}$ c. $\frac{3}{5}$ d. $\frac{1}{10}$ e. $\frac{7}{10}$ f. $\frac{2}{5}$