Respuesta :
Answer:
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
Step-by-step 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 = [tex]\frac{1}{2}[/tex]
Probability of getting a tail= [tex]\frac{1}{2}[/tex]
A coin is thrown 10 times and it come up heads 8 times.
p=(success of getting head)= [tex]\frac{1}{2}[/tex]
q=(failure of getting head)= [tex]\frac{1}{2}[/tex]
Formula of probability : [tex]^{n}C_r\times p^r\times q^{n-r}[/tex]
where n=total no of times=10 and r= no. of outcomes=8
[tex]P=^{10}C_8\times (\frac{1}{2})^r\times (\frac{1}{2})^{10-8}[/tex]
[tex]P=\frac{10\times 9\times 8!}{8!\times(10-8)!}\times (\frac{1}{2})^8\times (\frac{1}{2})^{2}[/tex]
[tex]P=\frac{10\times 9}{2\times1}\times (\frac{1}{2})^{10}[/tex]
[tex]P=45\times (\frac{1}{2})^{10}=45\times0.0009765625=0.0439453125[/tex]
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 : [tex]^{n}C_r\times p^r\times q^{n-r}[/tex]
where n=total no of times=10 and r= no. of outcomes=8
[tex]P=^{10}C_8\times (0.6)^r\times (0.4)^{10-8}[/tex]
[tex]P=\frac{10\times 9\times 8!}{8!\times(10-8)!}\times (0.6)^8\times (0.4)^2[/tex]
[tex]P=\frac{10\times 9}{2\times1}\times 0.01679616\times 0.16[/tex]
[tex]P=0.120932352[/tex]
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
[tex]P=\frac{0.04}{0.12}=0.33[/tex]
Last part of Question:
Assume at the outset there is an equal (.5,.5) prior probability of either coin.
Answer:
0.36
Step-by-step 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) = [tex]10C8 * 0.5^{2} *0.5^{2}[/tex]
P(E | A) =0.0440
P(E |B) = [tex]10C8 * 0.6^{2} *0.4^{2}[/tex]
P(E |B) = 0.1209
[tex]P(A | E) = \frac{P(E,A)}{P(E)} = \frac{P(E | A) P(A)}{P(E | A) P(A) + P(E | B) P(B)} \\P(A | E) = \frac{0.0440*0.5}{0.0440*0.5) + 0.1209*0.5} \\P(A | E) = 0.2665[/tex]
[tex]P(A | E) = \frac{P(E,B)}{P(E)} = \frac{P(E | B) P(B)}{P(E | A) P(A) + P(E | B) P(B)} \\P(A | E) = \frac{0.1209*0.5}{0.0440*0.5) + 0.1209*0.5} \\P(A | E) = 0.7334[/tex]
Probability of having the fair coin vs the probability of having the bent coin = 0.2665/0.7334
