Urban Workbench: You toss a fair coin three times: What is the probability of three heads, HHHHHH? What is the probability that you observe exactly one heads? Given that you have observed at least one heads, what is the probability that you observe at least two heads?

Thursday, August 19, 2021

You toss a fair coin three times:

What is the probability of three heads, $H H H$ ?
What is the probability that you observe exactly one heads?
Given that you have observed at least one heads, what is the probability that you observe at least two heads?

Solution
- We assume that the coin tosses are independent.
  1. $P (H H H) = P (H) \cdot P (H) \cdot P (H) = {0.5}^{3} = \frac{1}{8}$ .
  2. To find the probability of exactly one heads, we can write
    
    $P (One heads)$ $= P (H T T \cup T H T \cup T T H)$
    $= P (H T T) + P (T H T) + P (T T H)$
    $= \frac{1}{8} + \frac{1}{8} + \frac{1}{8}$
    $= \frac{3}{8}$ .
  3. Given that you have observed at least one heads, what is the probability that you observe at least two heads? Let $A_{1}$ be the event that you observe at least one heads, and $A_{2}$ be the event that you observe at least two heads. Then $A_{1} = S - {T T T}, and P (A_{1}) = \frac{7}{8};$ $A_{2} = {H H T, H T H, T H H, H H H}, and P (A_{2}) = \frac{4}{8} .$ Thus, we can write
    $P (A_{2} | A_{1})$ $= \frac{P (A_{2} \cap A_{1})}{P (A_{1})}$
    $= \frac{P (A_{2})}{P (A_{1})}$
    $= \frac{4}{8} . \frac{8}{7} = \frac{4}{7}$ .

Urban Workbench