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Consider two coins, $A$ and $B$. Let $p_A = 0.6$ be the probability that a flip of coin $A$ gives heads; let $p_B = 0.4$ be the corresponding probability for coin $B$.
Consider the following experiment:
$\bullet$ First we pick a coin randomly (with probability $1/2$), then flip it.
$\bullet$ Then:
– If the result of the first flip is heads, we then flip coin $A$ once.
– Otherwise, we flip coin $B$ once.
Thus in total we flip $2$ times. (The pick and the flips are jointly independent.) Here is a possible outcome of the experiment:
$\bullet$ 1. Pick $A$, flip $A$, get tails.
$\bullet$ 2. Flip $B$, get tails.
Let $X$ be the number of heads we get; it is a random variable.

(a) Compute the probability mass function of $X$, i.e., compute $P(X = x)$ for all possible values of $X$. Also find the cumulative distribution function $P(X \leq x)$ for $x \in \mathbb{R}$.
(b) Show that $P(\text{coin B is used in the second trial}) = 0.5$.
(c) Verify that $X \not∼ Binomial(2, 0.5)$ by comparing the probability mass functions.
(d) Consider the events $C_1 = \{\text{first trial gives heads}\}$, $C_2 = \{\text{second trial gives heads}\}$. Show that $C_1$ and $C_2$ are dependent.

$\mathbf{My~Attempts:}$
For part (a):
Since, we know we pick a coin randomly with probability of $1/2 = 0.5$.
So, this means $A$ and $B$ are equally likely which means $P(A) = P(B) = 0.5$.
Since, we know that the pick and the flips are jointly independent.
So, we have the probability of pick and flips is equal to the probability of pick times probability of flips.
Since, we know that if the result of the first flip is heads, we then flip coin $A$ once. Otherwise, we flip coin $B$ once.
So, this means no matter what coins we are using in the first flip. if the first flip is head then we must use coin $A$ again in the second flip and if first flip is tail then we must use coin $B$ in the second flip.
Therefore, $P(X=0) = 0.5 \times 0.4 \times 0.6 + 0.5 \times 0.6 \times 0.6 = 0.3$;
$P(X=1) = 0.5 \times 0.6 \times 0.4 \times 2 + 0.5 \times 0.4 \times 0.4 \times 2 = 0.4$;
$P(X=2) = 0.5 \times 0.6 \times 0.6 + 0.5 \times 0.4 \times 0.6 = 0.3$;
which $X=0$ with both tails in $2$ flips and $X=1$ as one tail and one head or vice versa in $2$ flips and $X = 2$ as both heads in $2$ flips.
Also, $P(X \leq x) = 0$ if $x < 0$; $P(X \leq x) = 0.3$ if $0 \leq x < 1$; $P(X \leq x) = 0.3 + 0.4 = 0.7$ if $1 \leq x < 2$ and $P(X \leq x) = 0.3 + 0.4 + 0.3 = 1$ if $x \geq 2$.

For part (b): Since, we know that if the result of the first flip is heads, we then flip coin $A$ once. Otherwise, we flip coin $B$ once.
So, this means no matter what coins we are using in the first flip. if the first flip is head then we must use coin $A$ again in the second flip and if first flip is tail then we must use coin $B$ in the second flip.
Also, we know that the pick and the flips are jointly independent.
So, $P(\text{coin B is used in the second trial})$ is equal to the probability of first flip is a tail and $A$ is used before the first flip plus the probability of first flip is a tail and $B$ is used before the first flip $= 0.5 \times 0.4 + 0.5 \times 0.6 = 0.5$

For part (c):
Since, by Binomial distribution, we know that $Binomial(2,0.5) = P(X=2) = \binom{2}{2} (0.5)^2 (1-0.5)^{2-2} = 0.25$
But, from part (a) we know that $P(X=2) = 0.3$
Therefore, $X \not\sim Binomial(2,0.5)$

For part (d):
notice that $P(C_1 \cap C_2)$ is equal to the probability of both flips are heads for coin $A$ plus the probability of both flip are heads for coin $B$ which $P(C_1 \cap C_2) = 0.5 \times 0.6 \times 0.6 + 0.5 \times 0.4 \times 0.6 = 0.3$
and $P(C_1)$ is equal to the probability of both flip are heads for both coin $A$ and $B$ plus first head and second tail for both coin $A$ and $B$ which $P(C_1) = 0.5 \times 0.6 \times 0.6 + 0.5 \times 0.6 \times 0.4 + 0.5 \times 0.4 \times 0.6 + 0.5 \times 0.4 \times 0.4 = 0.5$
and $P(C_2)$ is equal to the probability of both flip are heads for both coin $A$ and $B$ plus first tail and second head for both coin $A$ and $B$ which $P(C_2) = 0.5 \times 0.4 \times 0.4 + 0.5 \times 0.4 \times 0.6 + 0.5 \times 0.6 \times 0.4 + 0.5 \times 0.6 \times 0.6 = 0.5$
Then, we have $P(C_1) \cdot P(C_2) = 0.5 \times 0.5 = 0.25$
So, we have $P(C_1 \cap C_2) = 0.3 \neq 0.25 = P(C_1) \cdot P(C_2)$
Since, $P(C_1 \cap C_2) \neq P(C_1) \cdot P(C_2)$.
Therefore, $C_1$ and $C_2$ are not independent which means they are dependent.

$\mathbf{Is~that~my~attempt~of~part~(a)~to~(d)~all~correct~?}$
$\mathbf{And~are~there~anything~I~can~improve~or~add~?}$

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I think it is more formal to proceed as follows. Let $X_1$, $X_2$ be the outcomes of the first and second flips, where $X_i = 1$ if flip $i$ is heads. Then there are four possible outcomes: $$(X_1, X_2) \in \{(0,0), (0,1), (1,0), (1,1)\}.$$ The distribution of $X_1$ must be Bernoulli. Specifically, by the law of total probability, $$p_1 = \Pr[X_1 = 1] = \Pr[X_1 = 1 \mid A]\Pr[A] + \Pr[X_1 = 1 \mid B]\Pr[B] = p_A \cdot \frac{1}{2} + p_B \cdot \frac{1}{2} = 0.5,$$ hence $$X_1 \sim \operatorname{Bernoulli}(p_1 = 0.5).$$ Next, the outcome of the first flip determines the probability distribution of the second, specifically $$X_2 \mid X_1 \sim \operatorname{Bernoulli}(p_2)$$ where $$p_2 = p_A X_1 + p_B (1-X_1) = \begin{cases} p_A, & X_1 = 1 \\ p_B, & X_1 = 0. \end{cases}$$ It follows that $$\Pr[(X_1, X_2) = (0,0)] = \Pr[X_1 = 0]\Pr[X_2 = 0 \mid X_1 = 0] = (1-p_1)(1-p_B).$$ Similarly, $$\begin{align} \Pr[(X_1, X_2) = (0,1)] &= (1-p_1)p_B, \\ \Pr[(X_1, X_2) = (1,0)] &= p_1 (1-p_A) \\ \Pr[(X_1, X_2) = (1,1)] &= p_1 p_A. \end{align}$$ Therefore, if $X = X_1 + X_2$, we easily obtain $$\Pr[X = x] = \begin{cases} (1-p_1)(1-p_B), & x = 0 \\ (1-p_1)p_B + p_1(1-p_A), & x = 1 \\ p_1 p_A, & x = 2, \end{cases}$$ where upon substituting values we obtain $$\Pr[X = x] = \begin{cases} 0.3, & x = 0 \\ 0.4, & x = 1 \\ 0.3, & x = 2. \end{cases}$$ The CDF is straightforward from this.

For (b), the probability coin $B$ is used in the second trial is simply $\Pr[X_1 = 0] = 1 - p_1 = 0.5$.

For (c), observe if $Y \sim \operatorname{Binomial}(n = 2, p = 0.5)$, then $$\Pr[Y = 0] = \binom{2}{0} (0.5)^0 (1 - 0.5)^{2-0} = \frac{1}{4} \ne 0.3,$$ thus $X \not\sim Y$.

For (d), we proceed by simply observing $$\Pr[C_1 \cap C_2] = \Pr[(X_1, X_2) = (1,1)] = 0.3,$$ whereas $$\begin{align} \Pr[C_1] \Pr[C_2] &= \Pr[X_1 = 1]\Pr[X_2 = 1] \\ &= p_1 ((1-p_1)p_B + p_1 p_A) \\ &= (0.5)(0.5)(0.4 + 0.6) \\ &= 0.25 \ne 0.3. \end{align}$$

Your answers are correct and your reasoning is correct, but consist of more of a verbal exposition than a mathematical one. You can see how, although the above is virtually identical to your reasoning, it is mostly expressed in terms of mathematical equations rather than words. This has several advantages:

  • it reduces ambiguity;
  • the results readily generalize to other values for $p_A$ and $p_B$;
  • less effort is needed to follow the line of reasoning.
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