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15 Nov 2021, 02:50
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What is the probability of getting four tails and two heads when an unfair coin is tossed 6 times?
(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is \(\frac{2}{9}\).
(2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is \(\frac{4}{9}\).
(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is \(\frac{2}{9}\).
(2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is \(\frac{4}{9}\).
15 Nov 2021, 02:50
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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
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
General Discussion
21 May 2023, 20:45
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Bunuel
Hi Bunuel please let me know if my reasoning is okay.
(1) Since they are asking probability of getting TTTTHH.
I figured that I could square 2/9 to get the answer.
(2) With this information, I did not find a direct way to arrive at four tails and two heads and marked it insufficient.
Is that okay?
