Question

Suppose I have either a fair coin or a bent coin, and I don't know which. The bent coin has a 60% probability of coming up heads. I throw the coin ten times and it comes up heads 8 times. What is the probability I have the fair coin vs. the probability I have the bent coin?

Answer 1

Last part of Question:

Assume at the outset there is an equal (.5,.5) prior probability of either coin.

Answer:

0.36

Explanation:

Let P(A) = Probability of fair coin = 0.5

Let P(B) = Probability of bent coin = 0.5

The coin was thrown 10 times

The number of possible events, E = 10

P(E | A) =
`10C8 * 0.5^(2) *0.5^(2)`

P(E | A) =0.0440

P(E |B) =
`10C8 * 0.6^(2) *0.4^(2)`

P(E |B) = 0.1209

`P(A | E) = (P(E,A))/(P(E)) = (P(E | A) P(A))/(P(E | A) P(A) + P(E | B) P(B)) \\P(A | E) = (0.0440*0.5)/(0.0440*0.5) + 0.1209*0.5) \\P(A | E) = 0.2665`

`P(A | E) = (P(E,B))/(P(E)) = (P(E | B) P(B))/(P(E | A) P(A) + P(E | B) P(B)) \\P(A | E) = (0.1209*0.5)/(0.0440*0.5) + 0.1209*0.5) \\P(A | E) = 0.7334`

Probability of having the fair coin vs the probability of having the bent coin = 0.2665/0.7334

Answer 2

Probability of fair coin = 0.04

Probability of bent coin = 0.12

Probability of having the fair coin vs. the probability of having the bent coin is 0.33

Explanation:

Given : Two coins - one bent coin and one fair coin

In a fair coin,

Sample space-{H,T} = 2

Probability of getting a head =
`(1)/(2)`

Probability of getting a tail=
`(1)/(2)`

A coin is thrown 10 times and it come up heads 8 times.

p=(success of getting head)=
`(1)/(2)`

q=(failure of getting head)=
`(1)/(2)`

Formula of probability :
`^(n)C_r* p^r* q^(n-r)`

where n=total no of times=10 and r= no. of outcomes=8

`P=^(10)C_8* ((1)/(2))^r* ((1)/(2))^(10-8)`

`P=(10* 9* 8!)/(8!*(10-8)!)* ((1)/(2))^8* ((1)/(2))^(2)`

`P=(10* 9)/(2*1)* ((1)/(2))^(10)`

`P=45* ((1)/(2))^(10)=45*0.0009765625=0.0439453125`

Therefore, probability of head comes in fair coin = 0.04

In a bent coin,

Probability of getting a head = 60%=0.6

A coin is thrown 10 times and it come up heads 8 times.

p=(success of getting head)= 0.6

q=(failure of getting head)= 1-p=1-0.6=0.4

Formula of probability :
`^(n)C_r* p^r* q^(n-r)`

where n=total no of times=10 and r= no. of outcomes=8

`P=^(10)C_8* (0.6)^r* (0.4)^(10-8)`

`P=(10* 9* 8!)/(8!*(10-8)!)* (0.6)^8* (0.4)^2`

`P=(10* 9)/(2*1)* 0.01679616* 0.16`

`P=0.120932352`

Therefore, probability of head comes in bent coin = 0.12

So, probability of having the fair coin vs. the probability of having the bent coin is

`P=(0.04)/(0.12)=0.33`