Respuesta :
Answer:
The probability that the fair coin is the one chosen, given that the chosen coin lands on heads, is 0.64.
Step-by-step explanation:
Let H denote the event of coin toss getting a head. Let "Fair" and "Biased" refer to choices of the fair and the biased coins being tossed. P(A|B) refers to the conditional probability of event A given event B.
We are to determine the conditional probability P(Fair|H). This can be rewritten using the Bayes rule as:
[tex]P(Fair|H)=\frac{P(H|Fair)\cdot P(Fair)}{P(H)}[/tex]
in which P(Fair)=0.5 and P(H|Fair)=0.5. The only quantity that needs further thought is P(H), namely the a-priori probability of heads. This is a result of heads from a fair coin and a biased coin:
[tex]P(H) = P(H|Fair)P(Fair) + P(H|Biased)P(Biased)=\\=0.5\cdot0.5 + 0.28\cdot 0.5 = 0.39[/tex]
Using this value, the desired probability becomes:
[tex]P(Fair|H)=\frac{P(H|Fair)\cdot P(Fair)}{P(H)}=\frac{0.5*0.5}{0.39}=0.64[/tex]
The probability that the fair coin is the one chosen, given that the chosen coin lands on heads, is 0.64.
Using conditional probability, it is found that there is a 0.641 = 64.1% probability that the fair coin is the one chosen.
Conditional Probability
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
- P(B|A) is the probability of event B happening, given that A happened.
- [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
- P(A) is the probability of A happening.
In this problem:
- Event A: Lands on heads.
- Event B: Fair coin chosen.
The percentages of landing on heads are:
- 50% of 50%(fair coin).
- 28% of 50%(biased coin).
Hence:
[tex]P(A) = 0.5(0.5) + 0.28(0.5) = 0.39[/tex]
The probability of picking the fair coin and landing on heads is:
[tex]P(A \cap B) = 0.5(0.5) = 0.25[/tex]
Hence, the conditional probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.25}{0.39} = 0.641[/tex]
0.641 = 64.1% probability that the fair coin is the one chosen.
A similar problem is given at https://brainly.com/question/14398287
