What is the probability that at least one head is obtained from the unbiased coin?

Question

A loaded die is tossed once; if N is the result of the toss, then $P$($N = i$) = $p_i$ , $i = 1, ..., 6$. If $N = i$, an unbiased coin is tossed independently i times.

What is the probability that at least one head is obtained from the unbiased coin? Given that at least one head is obtained, what is the probability that N will be odd?

I know that a loaded die has probabilities 1/21, 2/21,3/21, 4/21, 5/21, 6/21 of showing 1,2,3,4,5,6. But, I am confused as to what $p_i$ is.

Answer 1

You just compute it. If the die shows $4$, you flip four coins and have $\frac {15}{16}$ chance of getting at least one head. Multiply that by $\frac 4{21}$ and add all the others.

Answer 2

I know that a loaded die has probabilities 1/21, 2/21,3/21, 4/21, 5/21, 6/21 of showing 1,2,3,4,5,6. But, I am confused as to what $p_i$ is.

Well now, $p_i$ is the probability that a roll of the biased dice shows face $i$.

It you have been told that these are the probabilities for your loaded die, then they are the six $(p_i)$ values.   However, if you have not been told so, then you cannot just presume it, and must leave the values as variables.

Now, you know that the probability for at least one head showing among $i$ tosses of an unbiased coin will be: $\mathsf P(H>0\mid N=i)=$ /////.

Where $H$ is of course the count for heads among the tosses of the unbiased coin.

Then, by the Law of Total Probability: $\mathsf P(H>0) = \sum_{i=1}^6 p_i \mathsf P(H>0\mid N=i) = $ /////.

Finally find $\mathsf P(N\in\{1,3,5\}\mid H>0)$ using the definition of conditional probability.

Divide and calculate.

Answer 3

1. If you flip an unbiased coin $k$ times, the chance of getting only tails is $\frac{1}{2^k}$, so the chance of getting heads at least once is $1-\frac{1}{2^k}$.
2. According to the way your die is biased, you have a $\frac{1}{21}$ chance of flipping one coin, $\frac{2}{21}$ of flipping two coins, etc.
3. Therefore, your probability of getting at least one heads is: $$\sum_{k=1}^6\frac{k}{21}(1-\frac{1}{2^k})=\frac{1}{21}\frac{1}{2}+\frac{2}{21}\frac{3}{4}+\frac{3}{21}\frac{7}{8}+\frac{4}{21}\frac{15}{16}+\frac{5}{21}\frac{31}{32}+\frac{6}{21}\frac{63}{64}=\frac{51}{56}$$