Launch of two coins with different probability of heads or tails

On the table, there are two coins, with head probability each $p_1=\frac{1}{5}$ and $p_2=\frac{1}{7}$.

I choose at random one of the two coins and repeatedly launch it. I get the first head on the 4th launch. What is the probability that I have in my hand the coin for which $p_1=\frac{1}{5}$ ?

I have thought that if I have launched the first coins, to get head at the 4th throw, the probability is $0{,}5\cdot (\frac{1}{5})^4$ because we have probability of $0{,}5$ for the chosen of the coins and every throw of the coins is independent of the others and have probability $\frac{1}{5}$

But I don't think this is the right solution

• The probability that you choose the first coin and get the first head on the $4$th launch is $0.5 \times (\frac45)^3\times \frac15$. Similarly the probability that you choose the second coin and get the first head on the $4$th launch is $0.5 \times (\frac67)^3\times \frac17$. You need to combine these results to get the answer Mar 30 '17 at 6:18
• Unfortunatly I can't understand the answer... And I can't figure out what class of problems is it. Mar 30 '17 at 14:56
• It is a conditional probability / Bayesian question Mar 30 '17 at 14:58
• So, I consider the events $E=$"launch of the first coin" and $F=$"head on the 4th launch". I need to find $P(E|F)=\frac{P(F|E)P(E)}{P(F|E)P(E)+P(F|E^C)P(E^C)}$ with $P(F|E)=\frac{4}{5}^3*\frac{1}{5}$, $P(F|E^C)=\frac{6}{7}^3*\frac{1}{7}$ and $P(E)=P(E^C)=0.5$. And at this I resolve the equation, right? Mar 31 '17 at 9:58
• Yes. You get the same result with $P(E\mid F)=\frac{P(E \cap F)}{P(F)}=\frac{P(E \cap F)}{P(E\cap F)+P(E^C\cap F)}$ Mar 31 '17 at 11:06

• $E=$"launch of the first coin"
• $F=$"head on the 4th launch"
I need to find $$P(E|F)=\frac{P(F|E)P(E)}{P(F|E)P(E)+P(F|E^C)P(E^C)}$$ with $$P(F|E)=(\frac{4}{5})^3∗\frac{1}{5}$$ and $$P(F|E^C)=(\frac{6}{7})^3∗\frac{1}{7}$$ I know $$P(E)=P(E^C)=0.5$$ So now I can replace with the number and find the answer
• Quite right. You might notice that since $P(E)$ and $P(E^C)$ are equal, your first formula simplifies to $$P(E|F)={P(F|E)\over P(F|E)+P(F|E^C)}$$ There's no getting around the rest of the arithmetic, though. Mar 31 '17 at 13:36