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Len [333]
3 years ago
6

Suppose thatAnn selects a ball by first picking one of two boxes at random and then selecting a ball from this box. The first bo

x contains three orange balls and four black balls, and the second box contains five orange balls and six black balls. What is the probability that Ann picked a ball from the second box if she has selected an orange ball?
Mathematics
1 answer:
Natalka [10]3 years ago
3 0

Answer:

35/68

Step-by-step explanation:

The first box contains 3 orange balls and 4 black balls.

The second box contains 5 orange balls and 6 black balls

Total number of boxes = 2

Let Pr(F) be the probability that the first box is picked.

Let Pr(S) be the probability that the second ball is picked

Let O be the orange ball selected

Pr(F) = 1/2

Pr(S) = 1/2

Pr(O|S) = n(O) / n(S)

= 5/ 6+5

5/11

Pr(O|F) = n(O) / n(F)

= 3/ 4+3

= 3/7

Pr(S|O) = [Pr( O|S).P(S)] / [Pr( O|S).P(S) + Pr( O|F).P(F)]

= (5/11*1/2) / (5/11*1/2) +(3/7*1/2(

= (5/22) / (5/22 + 3/14)

= (5/22) / (136/308)

= 35/68

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Answer:

a) (0.555, 0) and (6, 0)

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Set each term in the numerator and denominator equal to 0 and find r.

In the numerator:

r = 7/8, 5/9, or 6

In the denominator:

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Zeros in the numerator that aren't in the denominator are r-intercepts.

Zeros in the denominator that aren't in the numerator are vertical asymptotes.

Zeros in both the numerator and the denominator are holes.

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) Evaluate m(r) at r = 7/8.  To do that, first divide out the common term (-8r + 7) from the numerator and denominator.

m(r) = (-9r+5)² (r−6)² / ( (-5r+9)² (r+3)² )

m(⅞) = (-9×⅞+5)² (⅞−6)² / ( (-5×⅞+9)² (⅞+3)² )

m(⅞) = (-23/8)² (-41/8)² / ( (37/8)² (31/8)² )

m(⅞) = (-23)² (-41)² / ( (37)² (31)² )

m(⅞) = 0.676

The hole is at (0.875, 0.676).

d) Evaluate m(r) at r = 0.

m(0) = (-9×0+5)² (0−6)² / ( (-5×0+9)² (0+3)² )

m(0) = (5)² (-6)² / ( (9)² (3)² )

m(0) = 1.235

The m(r)-intercept is (0, 1.235).

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Step-by-step explanation:

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