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Virty [35]
4 years ago
10

17.5% as a fraction in simplest form?

Mathematics
1 answer:
Vsevolod [243]4 years ago
6 0
17.5\% is the same thing as 0.175. We can rewrite this as \frac{175}{1000}. Or course this is not simplified, so we can simplify to get \boxed{\frac{7}{40}}.
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150 students in a tenth grade high school class take a survey about which video game consoles they own 60 students answer that o
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The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

150 students in a tenth grade high school class take a survey about which video game consoles they own. 60  students answer that one of their consoles is a Playstation, 50 answer that one of their consoles is an Xbox. Out  of these, there are 20 who have both systems.

Let A be the event that a randomly selected student in the class has a Playstation and B be the event that the  student has an XBOX. Based on this information, answer the following questions.

a) What is P(A), the probability that a randomly selected student has a Playstation?

b) What is P(B), the probability that a randomly selected student has an XBOX?

c) What is P(A and B), the probability that a randomly selected student has a Playstation and an XBOX?

d) What is P(A | B), the conditional probability that a randomly selected student has a Playstation given that he  or she has an XBOX?

Answer:

a) P(A) = 2/5

b) P(B) = 1/3

c) P(A and B) = 2/15

d) P(A | B) = 2/5

Step-by-step explanation:

Total no. of students = 150

No. of students having playstation = 60

No. of students having xbox = 50

No. of students who have both playstation and xbox = 20

a) What is P(A), the probability that a randomly selected student has a Playstation?

P(A) = No. of students having playstation/Total no. of students

P(A) = 60/150

P(A) = 2/5

b) What is P(B), the probability that a randomly selected student has an XBOX?

P(B) = No. of students having xbox/Total no. of students

P(B) = 50/150

P(B) = 1/3

c) What is P(A and B), the probability that a randomly selected student has a Playstation and an XBOX?

The probability that a students has a Playstation and an Xbox is given by

P(A and B) = P(A)*P(B)

P(A and B) = (2/5)*(1/3)

P(A and B) = 2/15

d) What is P(A | B), the conditional probability that a randomly selected student has a Playstation given that he or she has an XBOX?

The conditional probability is given by

P(A | B) = P(A and B)/P(B)

P(A | B) = (2/15)/(1/3)

P(A | B) = 2/5

Alternatively:

P(A | B) = P(A∩B)/P(B)

Where P(A∩B) is given by

P(A∩B) = No. of students who have both playstation and xbox/Total no. of students

P(A∩B) = 20/150

P(A∩B) = 2/15

P(A | B) = P(A∩B)/P(B)

P(A | B) = (2/15)/(1/3)

P(A | B) = 2/5

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