# 5 coins. All biased, but differently biased.

So we have $5$ coins. $4$ of them have a $\frac 23$ chance of landing on heads. And the fifth has a $\frac 35$ chance of landing on heads. What is the chance that the $5$th coin lands on tails given that only one the $5$ coins that was flipped landed on tails.

I'm assuming you would do ${}_5C_1$, which gives you the number of ways where you can get $1$ tail. Then would you do a Bernoulli trial for both cases?

Can someone guide me through this?

• Since it's a conditional probability, you can calculate the probability that only the 5th coin was tails, and the probability that exactly one coin was tails, and then divide. Each probability can be done by hand (the latter is the sum of five individual probabilities). I don't think the 5C1 will help you here, since the coins aren't identical. Nov 10 '14 at 8:34
• You need to calculate the probabilities that one of the first four was tails and all others heads, and the probability that the fifth was tails and all others heads. Nov 10 '14 at 8:39
• @Henry so how would I calculate the prob that one of the first four was tails. Bernoulli? Nov 10 '14 at 8:51
• @GregMartin so would to find the prob that one of the first four was tails would I do: 1/3*1*1*1 for the first one being tail, 1/3*1/3*1*1 for the second being tail... and so on. Then for fifth being heads is 1/3*1/3*1/3*1/3*2/3. What would I do with these probs? Nov 10 '14 at 9:02
• For example, the probability that only the 5th coin is tails is $\frac23\cdot\frac23\cdot\frac23\cdot\frac23\cdot\frac25$; the probability that only the 1st coin is tails is $\frac13\cdot\frac23\cdot\frac23\cdot\frac23\cdot\frac35$. Do you see why? Nov 10 '14 at 17:10

Prescribe random variable $T_{i}$ by: $T_{i}=1$ if the $i$th coin lands on tails and $T_{i}=0$ otherwise.
Also let $S_{n}:=T_{1}+\cdots+T_{n}$ for $n=4,5$.
To be found is $P\left(T_{5}=1\mid S_{5}=1\right)$. This can be done on base of:
$$P\left(T_{5}=1\mid S_{5}=1\right)=\frac{P\left(T_{5}=1\wedge S_{5}=1\right)}{P\left(S_{5}=1\right)}=\frac{P\left(T_{5}=1\wedge S_{4}=0\right)}{P\left(S_{5}=1\right)}$$
• $P\left(T_{5}=1\wedge S_{4}=0\right)=P\left(T_{5}=1\right)P\left(S_{4}=0\right)$ (independence) and $P\left(S_{5}=1\right)=P\left(T_{5}=1\wedge S_{4}=0\right)+P\left(T_{5}=0\wedge S_{4}=1\right)=P\left(T_{5}=1\right)P\left(S_{4}=0\right)+P\left(T_{5}=0\right)P\left(S_{4}=1\right)$. Here $S_{4}$ is binomially distributed with $n=4$ and $p=\frac{1}{3}$. This is enough. Nov 10 '14 at 10:10