In an exam, there is a problem that 60% of students know the correct answer. However, thereis 15% chance that a student picked the wrong answer even if he/she knows the right answer andthere is also a 25% chance that a student does not know the right answer but guessed it correctly. If a student did get the problem right, what is the probability that this student really knows theanswer?

0.231

Step-by-step explanation:

Let the Probability of students that knew the correct answer be: P(A)

P(A) = 60% = 0.6

Let the Probability that the student picked the wrong answer even if he/she knows the right answer be: P(B)

P(B) = 15% =0.15

Let the Probability of the student that do not knew the correct answer Be P(C)

P(C) = 1 - P(A)

P(C) = 1 - 0.6

P(C) = 0.4

Let the Probability that the student does not know the right answer but guessed it correctly be: P(D)

P(D) = 25% = 0.25

Let the Probability that the student picked the right answer even if he/she knows the right answer be: P(E)

P(E) = 1 - P(B)

P(E) = 1 - 0.15

P(E) = 0.85

Probability that the student got the answer wrong = (0.60 X 0.15) + (0.40 X 0.75) = 0.39      

P( Student knew answer given he answered wrong) =

=

=

= 0.23076923077

= 0.231

2

step-by-2step explanation:

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15