HOT Competition 2 Sep/8AM: What is the probability of getting four tai

Official Solution:


What is the probability of getting four tails and two heads when an unfair coin is tossed 6 times?

Say the probability of tails is \(p\) and the probability of heads is \((1-p)\). The question asks to find the value of \(P(TTTTHH)=\frac{6!}{4!2!}*p^4*(1-p)^2\) (notice that we multiply by \(\frac{6!}{4!2!}\) because four tails and two heads can occur in \(\frac{6!}{4!2!}\) ways: TTTTHH, TTTHTH, TTHTTH, THTTTH, HTTTTH, ...

(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is \(\frac{2}{9}\).

\(P(TTH)=\frac{3!}{2!}*p^2*(1-p)=\frac{2}{9}\). From this we can find the value of \(p^2*(1-p)\), which when squared gives the value of \(p^4*(1-p)^2\), so we can find the value of \(P(TTTTHH)=\frac{6!}{4!2!}*p^4*(1-p)^2\). Sufficient.

(2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is \(\frac{4}{9}\).

\(P(TH)=2!*p*(1-p)=\frac{4}{9}\). From this we can find that the value of \(p\) is \(\frac{1}{3}\) or \(\frac{2}{3}\). These two values will give two different values of \(P(TTTTHH)=\frac{6!}{4!2!}*p^4*(1-p)^2\). Not sufficient.


Answer: A