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An unfair coin

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An unfair coin has the property that when flipped four times, it has the same probability of turning up 2 heads and 2 tails (in any order) as 3 heads and 1 tail (in any order). What is the probability of getting a head in any one flip?

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Let the probability of getting heads in a single flip be p.

The probability of getting 2 heads and 2 tails is (4 choose 2)*p^2*(1-p)^2

= 4!/2!2! * p^2*(1-p)^2

= 6 * p^2*(1-p)^2

The probability of getting 3 heads and 1 tails is (4 choose 3)*p^3*(1-p)

= 4!/3!1! * p^3*(1-p)

= 4 * p^3*(1-p)

Setting these equal gives

6 * p^2*(1-p)^2 = 4 * p^3*(1-p)

6 * (p-p^2)^2 = 4p^3-4p^4

6p^2 - 12p^3 + 6p^4 = 4p^3-4p^4

6p^2 - 16p^3 + 10p^4 = 0

5p^2 - 8p + 3 = 0 (divided by 2p^2 and reordered terms)

Applying the quadratic equation gives

p = (8 +/- sqrt( 64 - 4 * 5 * 3 ) ) / 10

p = (8 +/- sqrt( 64 - 60 ) ) / 10

p = (8 +/- sqrt( 4 ) ) / 10

p = (8 +/- 2 ) / 10

So p (which is the probability of getting a head in any one flip) is either 1 or 6/10.

Since this is a coin we're talking about, I doubt p=1.... so p = 6/10.

Here's a little verification:

If p is 1, the probability of both outcomes (2 head and 2 tails and 3 heads and 1 tails) is 0. Tails simply cannot come up.

If p is 6/10:

2h2t: 6 * p^2*(1-p)^2 = 6 * (.6)^2*(.4)^2 = 6^3 * 4^2 / 10^4 = .3456

3h1t: 4 * p^3*(1-p) = 4 * (.6)^3 * .4 = 6^3 * 4^2 / 10^4 =.3456

Edit: Since I divided by 2p^2 earlier.... it is possible p=0 too (but unlikely since it is a coin). In this case, like in the case of p=1, both the outcomes have probability 0.

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Posted · Report post

h = head probability

t = tail probability

4hhht = 6hhtt

2h = 3t

h = 1.5t

h + t = 1

1.5t + t = 1

2.5t = 1

t = 0.4

h = 0.6

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Posted · Report post

Both of you are absolutely correct... Witzar I liked the way with ease you worked it out...

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