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use for questions 9-12: a group of students were asked if they have at …

Question

use for questions 9-12: a group of students were asked if they have at least one sibling and at least one pet. of the students surveyed, 27 said they have a sibling and 24 students said they have a pet. of the students who said they had a sibling, 12 do not have pets.

  1. complete the venn diagram to the right to represent the results of this survey.

if one of these students is chosen at random, find each probability as a fraction in simplest form.

  1. ( p(\text{has a sibling and a pet}) )
  2. ( p(\text{does not have a pet}) )
  3. ( p(\text{has a pet but no sibling}) )

Explanation:

Step1: Find total number of students

First, we find the number of students in each section of the Venn diagram. For the "Sibling only" section, we know 12 students have a sibling but no pet. For the "Both" section, since 27 have a sibling and 12 have a sibling but no pet, we calculate \(27 - 12 = 15\) students have both a sibling and a pet. For the "Pet only" section, since 24 have a pet and 15 have both, we calculate \(24 - 15 = 9\) students have a pet but no sibling. Now, the total number of students is the sum of all sections: \(12+15 + 9=36\) (we can also check with the "no pet" count: 12 students have no pet, and 24 have a pet, so \(12 + 24=36\), which matches).

Step2: Calculate \(P(\text{has a sibling and a pet})\)

The probability is the number of students with both a sibling and a pet divided by the total number of students. The number of students with both is 15, and the total is 36. So we have \(\frac{15}{36}\), which simplifies by dividing numerator and denominator by 3: \(\frac{15\div3}{36\div3}=\frac{5}{12}\).

Step3: Calculate \(P(\text{does not have a pet})\)

The number of students who do not have a pet is 12 (from the "Sibling only" section, since that's the only section with no pet). So the probability is \(\frac{12}{36}\), which simplifies by dividing numerator and denominator by 12: \(\frac{12\div12}{36\div12}=\frac{1}{3}\).

Step4: Calculate \(P(\text{has a pet but no sibling})\)

The number of students with a pet but no sibling is 9. So the probability is \(\frac{9}{36}\), which simplifies by dividing numerator and denominator by 9: \(\frac{9\div9}{36\div9}=\frac{1}{4}\).

Answer:

  1. \(\boldsymbol{\frac{5}{12}}\)
  2. \(\boldsymbol{\frac{1}{3}}\)
  3. \(\boldsymbol{\frac{1}{4}}\)