Question 1209943
Let's break down this problem using Bayesian inference for each prior.

**1. Likelihood Function**

The likelihood function is the probability of observing 8 heads in 10 flips given a probability $p$ of heads. This follows a binomial distribution:

$L(p) = \binom{10}{8} p^8 (1-p)^2 = 45 p^8 (1-p)^2$

**2. Bayesian Inference**

Bayesian inference combines the likelihood function with a prior distribution to obtain a posterior distribution.

**a) No Prior Knowledge (Uniform Prior)**

* Prior: $P(p) = 1$ for $0 \le p \le 1$ (uniform distribution)
* Posterior: $P(p | \text{data}) \propto L(p) P(p) = 45 p^8 (1-p)^2$
* To find the estimated value of $p$, we need to find the maximum of the posterior. This is equivalent to maximizing the likelihood since the prior is uniform.
* To find the maximum, we take the derivative of $L(p)$ and set it to 0:
    * $L'(p) = 45 (8 p^7 (1-p)^2 - 2 p^8 (1-p)) = 0$
    * $8 (1-p) - 2 p = 0$
    * $8 - 8p - 2p = 0$
    * $10p = 8$
    * $p = 0.8$

* (1) Estimated $p = 0.8$
* (2) Yes, $p > 0.6$

**b) Coin is Likely Fair (Prior around 0.5)**

* Prior: A beta distribution centered around 0.5. Let's use $\text{Beta}(5, 5)$, for example.
    * $P(p) \propto p^4 (1-p)^4$
* Posterior: $P(p | \text{data}) \propto p^8 (1-p)^2 \cdot p^4 (1-p)^4 = p^{12} (1-p)^6$
* Maximize the posterior:
    * $12 (1-p) - 6 p = 0$
    * $12 - 12p - 6p = 0$
    * $18p = 12$
    * $p = 2/3 \approx 0.667$

* (1) Estimated $p \approx 0.667$
* (2) Yes, $p > 0.6$

**c) Coin is Likely Biased (Prior around 0 or 1)**

* Prior: A beta distribution centered around 0 or 1. Let's use $\text{Beta}(8, 2)$, for example.
    * $P(p) \propto p^7 (1-p)^1$
* Posterior: $P(p | \text{data}) \propto p^8 (1-p)^2 \cdot p^7 (1-p)^1 = p^{15} (1-p)^3$
* Maximize the posterior:
    * $15 (1-p) - 3 p = 0$
    * $15 - 15p - 3p = 0$
    * $18p = 15$
    * $p = 5/6 \approx 0.833$

* (1) Estimated $p \approx 0.833$
* (2) Yes, $p > 0.6$

**d) 𝑝 can only take values 0.2, 0.7, or 0.9**

* Prior: We need to assign probabilities to each value. Let's assume equal prior probabilities: $P(0.2) = P(0.7) = P(0.9) = 1/3$.
* Likelihoods:
    * $L(0.2) = 45 (0.2)^8 (0.8)^2 \approx 0.000046$
    * $L(0.7) = 45 (0.7)^8 (0.3)^2 \approx 0.19$
    * $L(0.9) = 45 (0.9)^8 (0.1)^2 \approx 0.196$
* Posterior (proportional to likelihood since prior is uniform):
    * $P(0.2 | \text{data}) \propto 0.000046$
    * $P(0.7 | \text{data}) \propto 0.19$
    * $P(0.9 | \text{data}) \propto 0.196$
* The highest posterior is for $p = 0.9$.

* (1) Estimated $p = 0.9$
* (2) Yes, $p > 0.6$

**e) 0.4 ≤ 𝑝 ≤ 0.9**

* Prior: A uniform distribution on the interval $[0.4, 0.9]$.
    * $P(p) \propto 1$ for $0.4 \le p \le 0.9$, and $0$ elsewhere.
* Posterior: $P(p | \text{data}) \propto p^8 (1-p)^2$ for $0.4 \le p \le 0.9$, and $0$ elsewhere.
* The maximum of $p^8 (1-p)^2$ occurs at $p = 0.8$, which is within the range $[0.4, 0.9]$.

* (1) Estimated $p = 0.8$
* (2) Yes, $p > 0.6$

**Summary**

In all cases, the estimated value of $p$ is greater than 0.6. The exact value of $p$ varies depending on the prior information, but it is always in the range of 0.667 to 0.9.