1. ## Probability question

a) = 1/2
b) = 1/2
c)= P|AnB|/P|S| = 3 /2^3 = 3/8 AnB= 3
Would you be able to consider part c) in positions for AnB e.g. position 1 _ position 2 _ etc.. or no?

2. ## Re: Probability question

I don't understand what you've written for (c)

\begin{align*} &P[A \cap B] = \\ \\ &P[\text{1 tail}] = \\ \\ &\dbinom{3}{1} \left(\dfrac 1 2\right)^3 = \dfrac 3 8 \end{align*}

If $A$ and $B$ are independent events then

$P[A \cap B] = P[A]P[B]$

from above

$P[A\cap B] = \dfrac 3 8 \neq \dfrac 1 4 = \left(\dfrac 1 2 \right )^2 = P[A]P[B]$

thus $A$ and $B$ are not independent.