Two dice with a difference ... They appear identical, but one die is fair and the other is loaded so that the probability of throwing a 6 is 0.25. Bev picks a die at random and throws it, and it shows a 6. Find the probability that she picked the biased die.

(Hint: Let A be the event 'the die chosen is the biased one' and B be 'a die chosen at random shows a 6 when it is thrown'. Find $P(A|B).)

I know $P(A) = 1/2$, $P(B) = (1/2 \times 0.25) + (1/2 \times 1/6) = 5/24$

Don't know what $P(A|B)$ is and what to do do next.


The basic formula that defines the conditional probability $\Pr(A|B)$ is $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.\tag{1}$$ You already computed $\Pr(B)$.

As to $\Pr(A\cap B)$, the probability that you picked the biased die and threw a $6$, you also already computed it as part of the calculation of $\Pr(B)$: It is $(1/2)(0.25)$.

So all that is left to do is the division asked for by Formula (1).

  • $\begingroup$ Ahhh thank you, Andre! I was twisting myself up just thinking about this thing. It's really simple with your explanation. Thaaaank yaaaaaaaaw! $\endgroup$ – dataquest Jun 4 '14 at 21:04
  • $\begingroup$ You are welcome. Formula (1) is very basic, and comes up explicitly or implicitly whenever one is dealing with conditional probabilities. $\endgroup$ – André Nicolas Jun 4 '14 at 21:06
  • $\begingroup$ Final answer without rounding:p(A/B) = 0.125/0.20833333 = 0.6000000096000002 $\endgroup$ – dataquest Jun 4 '14 at 21:07
  • $\begingroup$ The exact answer is $\frac{3}{5}$, which is $0.6$. The extra digits you got come from truncating. Your calculator or calculator program would probably have come up with $0.6$ if you did not rekey partial computations, and used instead the "memory" feature. $\endgroup$ – André Nicolas Jun 4 '14 at 21:15

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