8/9
This problem involves calculating a conditional probability, specifically finding the probability that the selected coin is double-headed given that it landed heads three times in a row. We can solve this using Bayes' Theorem.
Defining Events
- Let DH be the event that a double-headed coin is selected.
- Let F be the event that a fair coin is selected.
- Let E be the event that the coin is flipped 3 times and results in 3 heads (HHH).
Prior Probabilities
There are 8 coins in total (4 fair + 4 double-headed).
- The probability of selecting a double-headed coin is $P(\text{DH}) = \frac{4}{8} = \frac{1}{2}$.
- The probability of selecting a fair coin is $P(\text{F}) = \frac{4}{8} = \frac{1}{2}$.
Likelihood of the Evidence
We need to find the probability of observing 3 heads (E) given the type of coin selected.
- Probability of E given a double-headed coin (DH): A double-headed coin always lands heads. So, the probability of getting 3 heads in a row is: $P(\text{E} | \text{DH}) = 1 \times 1 \times 1 = 1$
- Probability of E given a fair coin (F): A fair coin has a $1/2$ probability of landing heads on each flip. So, the probability of getting 3 heads in a row is: $P(\text{E} | \text{F}) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$
Calculating the Total Probability of the Evidence (P(E))
We use the law of total probability to find the overall probability of observing 3 heads:
$P(\text{E}) = P(\text{E} | \text{DH}) P(\text{DH}) + P(\text{E} | \text{F}) P(\text{F})$ $P(\text{E}) = \left(1 \times \frac{1}{2}\right) + \left(\frac{1}{8} \times \frac{1}{2}\right)$ $P(\text{E}) = \frac{1}{2} + \frac{1}{16}$ $P(\text{E}) = \frac{8}{16} + \frac{1}{16} = \frac{9}{16}$Applying Bayes' Theorem
Now we apply Bayes' Theorem to find the probability that the coin was double-headed given the evidence of 3 heads:
$P(\text{DH} | \text{E}) = \frac{P(\text{E} | \text{DH}) P(\text{DH})}{P(\text{E})}$ $P(\text{DH} | \text{E}) = \frac{1 \times \frac{1}{2}}{\frac{9}{16}}$ $P(\text{DH} | \text{E}) = \frac{\frac{1}{2}}{\frac{9}{16}}$ $P(\text{DH} | \text{E}) = \frac{1}{2} \times \frac{16}{9}$ $P(\text{DH} | \text{E}) = \frac{16}{18}$ $P(\text{DH} | \text{E}) = \frac{8}{9}$Therefore, the probability that the coin drawn was double-headed, given that it landed heads 3 times, is $8/9$.
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