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fiasKO [112]
3 years ago
5

A teacher has found that the probability that a student studies for a test is 0.610.61​, and the probability that a student gets

a good grade on a test is 0.790.79​, and the probability that both occur is 0.560.56. a. Are these events​ independent? b. Given that a student​ studies, find the probability that the student gets a good grade. c. Given that a student gets a good​ grade, find the probability that the student studied.
Mathematics
1 answer:
Ann [662]3 years ago
7 0

Answer with step-by-step explanation:

Let

A=Student studies for a test

B=Student gets good grade on a test

The probability that a student studies  for  a test=P(A)=0.61

The probability that a student gets   a good grade on a test=P(B)=0.79

The probability that both occur=P(A\cap B)=0.56

a.We have to find the events are independent

We know that if two events A and B are independent then

P(A)\cdot P(B)=P(A\cap B)

P(A)\cdot P(B)=0.61\times 0.79=0.4819

P(A\cap B)\neq P(A)\cdot P(B)

Hence, given events are not independent.

b.We have to find P(B/A)

P(B/A)=\frac{P(A\cap B)}{P(A)}

P(B/A)=\frac{0.56}{0.61}=0.92

c. We have to find P(A/B)

P(A/B)=\frac{P(A\cap B}{P(B)}

P(A/B)=\frac{0.56}{0.79}=0.71

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H&S = 1,014 M; and S&B = 854m. This is resolved using the Sine Theorem.

<h3>What is the calculation backing the above solution?</h3>

Notice that the Angles of a Triangle add up to 180° That is

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1 year ago
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2 years ago
A six-sided die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on th
agasfer [191]

Answer:

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

Step-by-step explanation:

Given that; the probability of each face turning up is proportional to the number of dots on that face

P(1) = 1×P(1)

P(2) = 2×P(1)

P(3) = 3×P(1)

P(4) = 4×P(1)

P(5) = 5×P(1)

P(6) = 6×P(1)

P(T) = 21×P(1)

Where;

P(x) is the probability of getting number x on the dice.

P(T) is the total probability of obtaining any number

N(x) is the number of possible number x in terms of the distribution function.

P(x) = N(x)/N(T) ....1

And since P(T) is constant, and P(T) is proportional to N(T) then,

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The probability of getting either a 5 or a 2 in one throw

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Substituting the values of each probability;

P(2U5) = (2P(1) + 5P(1))/21P(1)

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P(1) cancel out, to give;

P(2U5) = 7/21 = 1/3

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