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Ganezh [65]
2 years ago
13

There is a 30 percent chance that A can fix her busted computer. If A cannot, then there is a 40 percent chance that her friend

B can fix it.
a) Find the probability it will be fixed by either A or B.
b) If it is fixed, what is the probability it will be fixed by B.
Mathematics
1 answer:
Solnce55 [7]2 years ago
8 0

Answer:

(a) 0.70

(b) 0.40

Step-by-step explanation:

Mutually-exclusive events are those events which cannot occur together.

Consider that events X and Y are mutually exclusive.

P (X and Y) = 0

Here the two events can be defined as follows:

A = A can fix the busted computer

B = B can fix the busted computer

The information provided are as follows:

P (A) = 0.30

P (B) = 0.40

If A cannot, then there is a chance that her friend B can fix it.

The above statement suggest that the events A and B are mutually exclusive, i.e. if A can fix the computer then B does not have to and if cannot then only B will fix it.

That is, P (A and B) = 0.

(a)

Compute the probability it will be fixed by either A or B as follows:

P (A or B) = P (A) + P (B) - P (A and B)

                = 0.30 + 0.40 - 0

                = 0.70

Thus, the probability it will be fixed by either A or B is 0.70.

(b)

Compute the probability that if it is fixed it will be fixed by B as follows:

P (not A and B) = P (B) - P (A and B)

                         = 0.40 - 0

                         = 0.40

Thus, the probability that if it is fixed it will be fixed by B is 0.40.

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2 is prime, so no factors.

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Solve with cramer's rule x+2y+3z=11, 2x+y+2z=10, 3x+2y+z=9
nalin [4]

Answer:

x = 2 , y = 0 , z = 3

Step-by-step explanation:

Cramer's rule is a rule through which we can find the solution of linear equation.

we have the three linear equations as

x+2y+3z=11

2x+y+2z=10

3x+2y+z=9

AX=B  

A: coefficient matrix

X= unknown vectors(x,y,z)

D = values of the linear equation (11 , 10 , 9)

now we find the determinant of the given linear equation

determinant of the matrix will be

A = \left[\begin{array}{ccc}1&2&3\\2&1&2\\3&2&1\end{array}\right]  = 1(1-4) - 2(2-6) + 3(4 - 3)

                    = 1(-3) - 2(-4) + 3(1)

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also D\neq 0

so the determinant is Non zero we can apply Cramer's rule

we will be replacing the first column of the coefficient matrix A with the values of D

by replacing the first column we will get the value of the variable 'x'

Dx =  \left[\begin{array}{ccc}11&2&3\\10&1&2\\9&2&1\end{array}\right]   = 11(1-4) -2(10-18) + 3(20-9) = -33+16+33 = 16

x = \frac{Dx}{D}  = \frac{16}{8} = 2

similarly

Dy = \left[\begin{array}{ccc}1&11&3\\2&10&2\\3&9&1\end{array}\right] = 1(10-18) -11(2-6) + 3(18 -30) = -8 +44 -36 = 0

y = \frac{Dy}{D} = 0

Dz= \left[\begin{array}{ccc}1&2&11\\2&1&10\\3&2&9\end{array}\right] = 1(9 - 20) -2(18 - 30) + 11(4 -3) = -11 +24 +11 = 24

z = \frac{Dz}{D} = \frac{24}{8} = 3

so we have the solution as

x = 2 , y = 0 , z = 3

Therefore the solution for the given linear equations is (2,0,3).

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