1
$\begingroup$

You are given with three dices, two normal and one fake. The fake dice possess two "5" faces and no "4" faces.

You choose one dice out of the three and roll it twice.

  1. What is the probability that in the first two rolls, the dice will fall on "5" ?

  2. If one knows that in the second roll the dice fell on "5", what is the probability that in the first roll the dice fell on a "4"?

  3. If one knows the chosen dice fell twice on "5", and then we choose one dice out of the other two and roll it. What is the probability the dice will fall on a "5"?

In question 1, my calculation was: $\frac{2}{3} \frac{1}{6} \frac{1}{6} + \frac{1}{3} \frac{1}{3} \frac{1}{3} $ . The first term corresponds to choosing a normal dice and the second term corresponds to choosing a fake dice. The answer is $\frac{1}{18}$ (I think that the two consecutive rolls are independent, so I can just multiply the probabilities for intersection).

In question 2, my calculation was $\frac{2}{3} \frac{1}{6^2} = \frac{1}{54} $ .

My problem is with the third part of the question. How can I calculate this part using conditional probability ? The denominator should be $\frac{1}{18} $ , but when calculating the numerator I get a number bigger than $\frac{1}{18}$ . Will you please help me understand the calculation ?

Thanks in advance

$\endgroup$

2 Answers 2

0
$\begingroup$

The answer to your first question is correct.

For the second question you should use that for the probability of an event $A$ given the event B can be calculated by: $$\mathbb{P}(A\ |\ B) = \frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}.$$ If you now define $A:=$ "the first roll gives a 4" and $B:=$ "the second roll gives a five" you find your answer. If I'm correct this is $1/12$.

COMPUTATION: $$\mathbb{P}(A\ |\ B) = \frac{\frac{2}{3}\frac{1}{6}\frac{1}{6}+\frac{1}{3}0\frac{1}{3}}{\frac{2}{3}\frac{1}{6}+\frac{1}{3}\frac{1}{3}}=\frac{1}{12}.$$

Similar for the third question, but now $A:=$ "the third roll gives a 5" and $B:=$ "the first two rolls give a 5". $\mathbb{P}(B)$ was already calculated in the first question.

COMPUTATION:

$$\mathbb{P}(A\ |\ B) = \frac{\frac{2}{3}\left(\frac{1}{6}\right)^2\left(\frac{1}{2}\frac{1}{6}+\frac{1}{2}\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{3}\right)^2\left(1\cdot\frac{1}{6}\right)}{\frac{1}{18}} = \frac{7}{36}.$$

In the numerator de terms between brackets give the probability of choosing a third dice and rolling a 5, the squared numbers give the probability of throwing a 5 twice with the first dice and the remaining numbers give the probability of choosing the first dice.

Hope this is clear and the arithmetics are correct.

$\endgroup$
0
0
$\begingroup$

Im not so sure but maybe:

  • a) The distributed probability for 2 consecutive 5 (choosing one dice between 3 where 2 are normal and 1 is fake) is $\frac{1}{6^2}\frac{2}{3}$ on normal dice and $\frac{1}{3^2}\frac{1}{3}$ on the fake dice, so you have $P_n=\frac{\frac{1}{6^2}\frac{2}{3}}{\frac{1}{6^2}\frac{2}{3}+\frac{1}{3^2}\frac{1}{3}}=\frac{1}{3}$ probability that the rolled dice was normal instead of fake and $P_f=1-P_n=\frac{2}{3}$ probability that the rolled dice was fake instead of normal (where n is for "normal" and f is for "fake").

  • b) If the rolled dice was the fake then the probability to take a 5 choosing one of the others dice is $P_p(5)=\frac{1}{6}$, because these two have the same probability on 5 (p is of "pure").

  • c) If the rolled dice was a normal one then the probability to take a 5 choosing one of the others is $\frac{1}{6}\ or\ \frac{1}{3}\ i.e.\ P_m(5)=\frac{1}{3}\frac{1}{2}+\frac{1}{6}\frac{1}{2}=\frac{1}{4}$ (the m is for "mixed").

  • d) Combining you knowledge from a) then the probability to take a 5 in the described circumstances is

$P(5)=P_p*P_f+P_m*P_n=\frac{1}{6}\frac{2}{3}+\frac{1}{4}\frac{1}{3}=\frac{7}{36}\approx 19.4\%$

I hope this maybe correct/useful for you.

EDIT: forget to distribute the probabilities of the first roll, now seems correct.

$\endgroup$
1
  • $\begingroup$ Thanks a lot !!! Unfortunately I can't give an "answer" to both of you, but you also helped me a lot ! Thanks for that ! $\endgroup$ Commented Sep 6, 2014 at 6:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .