If you randomly pick a coin from a box that contains $90\%$ fair coins and $10\%$ loaded coins, (a loaded coin gives heads $90\%$ of the time), toss it $5$ times and get all heads. What is the probability that this is a loaded coin?

$$n = 5, x = 5, p = 0.1(0.9)^5$$

Where $n$ is the number of flips, $x$ is number of heads, and $p$ is the probability of getting heads. so $$( 0.1 (0.9)^5 ) + (0.9 (0.5)^5) = 0.087$$

Can anyone see if the answer I provided is correct or not !

  • $\begingroup$ It looks like you asked this question before and we told you what the right formula is. It looks like the parentheses are messed up .It might be right if the second multiplication sign was a plus sign. There should not be a term involving 0.9 raised to the 10th power which it seems you have. $\endgroup$ Commented Oct 9, 2017 at 1:47
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    $\begingroup$ This is a duplicate of his earlier question but not of Checking whether a coin is fair. I vote to leave it open. $\endgroup$ Commented Oct 9, 2017 at 3:51

1 Answer 1



Try using these formula.

From Bayes Theorem

$$P(\text{loaded }| 5\text{ heads}) = P(5\text{ heads|loaded}) \frac{P(\text{loaded})}{P(\text{5 heads)}}$$ where by law of total probability: $$P(5 \text{ heads}) =P(5\text{ heads|loaded}) P(\text{loaded})+ P(5\text{ heads|fair}) P(\text{fair})$$

  • $\begingroup$ That is what we suggested in the previous post. I gave you +1 anyway. $\endgroup$ Commented Oct 9, 2017 at 1:49
  • $\begingroup$ I made an edit to my question see if now my answer is correct, then I will approve your posted answer :) $\endgroup$ Commented Oct 9, 2017 at 2:24
  • $\begingroup$ You have the formula right now. $\endgroup$ Commented Oct 9, 2017 at 2:26
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    $\begingroup$ You might want to state what do you think the values for $P(\text{fair})$, $P(5 \text{ heads}|\text{fair})$, $P(\text{loaded})$, $P(5 \text{ heads}|\text{loaded})$ are explicitly before you apply the formula. $\endgroup$ Commented Oct 9, 2017 at 2:31
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    $\begingroup$ I can see that you mean $0.5^5$. Now you have all the ingredient, you just have to sustitute them to get the desired quantity. $\endgroup$ Commented Oct 9, 2017 at 2:56

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