Cheers, I am given an unfair coin for which I know that if we toss it $10$ times, the probability of getting $5$ heads is double the probability of getting $4$ heads. I am then asked to find the probability of at least one head in $5$ tosses.
My try:
Let success be getting heads with a probability of $p = P(S)$
Let failure be getting tails with a probability of $1-p = P(F)$
If we assume that the tosses are independent, we can use the Binomial Distribution to compute the probability, so if we toss the coin $10$ times, and we get $5$ heads (successes), then: $P(5) = \begin{pmatrix}10 \\ 5\end{pmatrix}p^5(1-p)^5$ and if out of 10 tosses we get $4$ heads (successes) then $P(4) = \begin{pmatrix}10 \\ 4\end{pmatrix}p^4(1-p)^6$. From that we know, $P(5) = 2P(4) \rightarrow \begin{cases} p = 0 \\p= \frac{5}{8} \\ p = 1\end{cases}$ so I choose $p=\frac{5}{8}$ and then for at least one heads in 5 tosses: $P(X \geq 1) = P(X = 1) + P(X=2) + P(X = 3) + P(X=4) + P(X=5) = 1 – P(X = 0)= 1- \begin{pmatrix}5 \\ 0\end{pmatrix}(\frac{5}{8})^0(1-\frac{5}{8})^5 = 0.99$.
Would something like this work? Thanks for your help and your time =)
Best Answer
Yes, your reasoning is correct. There isn't much else to say; your work is complete as written.
As an exercise, how would you generalize the question? For instance, suppose we are told that in $n$ tosses, the probability of getting $x+1$ heads is $k$ times the probability of getting $x$ heads. What is $p$ as a function of $x$, $k$, and $n$? What is the probability that there is at least one heads in $m$ tosses?