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I'm trying to do a question in probability:

''we flip three coins''

''What is the probability that the second coin landed tails, given that two coins (exactly) landed head?''

I have set out the sample space - S= {HHH, HHT, THH, HTH, HTT, TTH, THT, TTT} E1 - tails on the second coin - {HTH} E2 - outcomes feat. 2 heads exactly - {HHT, HTH, THH}

P(E1 + E2) = 1/8

P(E2) = 3/8

P(E1/E2) = P(E1+E2)/P(E2) = [(1/8)]/[(3/8)] = 1/3

? Is this correct or will I have to use Bayes' theorem for this?

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    $\begingroup$ No that's correct. Think of the conditioning like restricting your sample space to only {HHT,HTH,THH}. Then you see that only one of the three gives the desired event. But equally going through the motions with conditional probability rule works. $\endgroup$
    – Michael
    Commented Jan 22, 2015 at 21:03

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$\color{green}{\checkmark}$ The calculations using Conditional Probability are correct. $\mathsf P(A\cap B)=\mathsf P(A)\mathsf P(B\mid A)$

  • To use Bayes' Theorem you'd need the other conditional probability. $\mathsf P(A\mid B)\;\mathsf P(B)= \mathsf P(B\mid A)\;\mathsf P(A)$

Your labeling of the events is a bit misleading.

You have:

the sample space - S= {HHH, HHT, THH, HTH, HTT, TTH, THT, TTT}

E1 - tails on the second coin - {HTH}

No, tails on the second coin is: {HTH, HTT, TTH, TTT}

E2 - outcomes feat. 2 heads exactly - {HHT, HTH, THH}

And that leads to: E1 $\cap$ E2 : {HTH} , tails on the second coin and exactly two heads.

Which, despite the labelling, you then have used correctly in the conditional probability calculation.

$$\begin{align} \mathsf P(E_1\mid E_2) & = \frac{\mathsf P(E_1\cap E_2)}{\mathsf P(E_2)} \\ & = \frac{1/8}{3/8} \\ & = \tfrac{1}{3} \end{align}$$

Showing that it was just an error in labelling not understanding. But still, try to avoid doing that on an exam.

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  • $\begingroup$ I'm confused with the interchange between Bayes' theorem and the 'typical' conditional probability formula - is the 'typical' formula accredited to Baye? $\endgroup$
    – Edward
    Commented Jan 22, 2015 at 21:52
  • $\begingroup$ Well, fair enough. Conditional Probability is defined as, $\;\mathsf P(A\mid B) \mathop{:=} \frac{\mathsf P(A\cap B)}{\mathsf P(B)}\;$, and Bayes' Law is derived from that: $\;\mathsf P(A\mid B) = \frac{\mathsf P(B\mid A)\mathsf P(A)}{\mathsf P(B)}\;$. $\endgroup$ Commented Jan 22, 2015 at 22:59

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