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Pie
3 years ago
5

There are two coins, one is fair and one is biased. The biased coin has a probability of landing on heads equal to 4/5. One of t

he coins is chosen at random (50-50), and is flipped repeatedly until it lands on tail. If it landed on heads 4 times before landing on tails, what is the posterior probability that coin chosen was biased?
Mathematics
1 answer:
den301095 [7]3 years ago
3 0

Answer:

72.4%

Step-by-step explanation:

The probability of A occurring given that B occurs = the probability of both A and B / the probability of B

P(A|B) = P(A∩B) / P(B)

This can be rearranged as:

P(A∩B) = P(B) P(A|B)

In this case:

A = biased coin is chosen

~A = fair coin is chosen

B = 4 heads then 1 tail

First, let's find P(A∩B).

P(A∩B) = P(B) P(A|B)

P(A∩B) = ½ × ₅C₄ (⅘)⁴ (⅕)¹

P(A∩B) = 0.2048

Next, find P(~A∩B).

P(~A∩B) = P(B) P(~A|B)

P(~A∩B) = ½ × ₅C₄ (½)⁴ (½)¹

P(~A∩B) = 0.078125

Therefore, the probability that the coin is biased is:

P = P(A∩B) / (P(A∩B) + P(~A∩B))

P = 0.2048 / (0.2048 + 0.078125)

P = 0.723866749

The probability is approximately 72.4%.

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(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

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We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

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... 4 = y . . . . . . . . cube root

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2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

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... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

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