4
$\begingroup$

A game uses an unbiased die with faces numbered 1 to 6. The die is thrown once. If it shows 4 or 5 or 6 then this number is the final score. If it shows 1 or 2 or 3 then die is thrown again and the final score is the sum of the numbers shown on both throws.

i. Find the probability that the final score is 4.

ii. Given the die is thrown only once, find the probability that the final score is 4.

iii. Given the die is thrown twice, find the probability that the final score is 4.

I have managed to solve part i and ii and would share the solution below. I am unable to work out part iii and would appreciate help.

My solution:

i. $P(4) + P(1,3) + P(2,2) + P(3,1) = \dfrac{1}{6} + \dfrac{1}{36} \cdot 3 = \dfrac{1}{4}$

ii. We know that the die is thrown only once and that can happen only when the score is 4, 5 or 6. The sample space is 3. So $P(4)$ now is $1/3$.

iii. My guess: $P(1,3) + P(2,2) + P(3,1) = 1/12$ --> Apparently this answer is incorrect and I really can't work out why this is wrong and what the correct approach would be.

Reference: OCR Jan 2009 Probability & Statistics 1 (4732)

$\endgroup$
  • $\begingroup$ an interesting variation - throw dice until score is $\ge 4$. $\endgroup$ – JMP May 24 '15 at 0:45
4
$\begingroup$

Your answers and reasoning are correct for parts 1 and 2.

Given that the die is thrown twice implies that the first die throw is either a 1 a 2 or a 3.

In each of these cases there is only one result for the second die that will cause your final total score to be 4 (if the first die was a 1, second die must be a 3; if first die was a 2, second die must be a 2; etc)

The probability of hitting the required number is the same in each of the cases, namely $\frac{1}{6}$. Thus, the probability is in fact $\frac{1}{6}$.


Approaching this from a more formal viewpoint: let our sample space be all ways of throwing two dice consecutively. Let $A$ represent the event that we get a score of four (where the second roll is ignored if first roll is high enough), and let $B$ represent the event that both rolls are counted (i.e. first roll is a 1,2, or 3).

Your problem asks to find $Pr(A|B)$

By definition: $Pr(A|B) := \frac{Pr(A\cap B)}{Pr(B)}$

As you correctly noted, $A\cap B = \{(1,3),(2,2),(3,1)\}$ and so $Pr(A\cap B) = \frac{3}{36}$.

And note: $Pr(B) = \frac{18}{36} = \frac{3}{6} = \frac{1}{2}$ (as there are 3 possibilities out of 6 for the first die roll to trigger the necessity of the second die roll).

So, $Pr(A|B) = \frac{Pr(A\cap B)}{Pr(B)} = \frac{3/36}{1/2} = \frac{2}{12} = \frac{1}{6}$

$\endgroup$
  • $\begingroup$ but why is my solution wrong? $\endgroup$ – student May 24 '15 at 0:37
  • $\begingroup$ You have to divide by the probability that two throws are needed. $\endgroup$ – Brian Tung May 24 '15 at 0:38
  • $\begingroup$ @student added a more formal approach. You essentially forgot to divide by $Pr(B)$ $\endgroup$ – JMoravitz May 24 '15 at 0:38
  • $\begingroup$ also considering your case, the required number for second throw can either be 1, 2 or 3. So we should really be adding P(1), P(2) and P(3). This would give 0.5? $\endgroup$ – student May 24 '15 at 0:40
  • $\begingroup$ @student only one of those numbers would be a success at a given time. Informally, you would multiply each by the probability that you are actually in that respective case. So, essentially (trying to match your notation) $P(\text{first is}~1)P(\text{second is}~3) + P(\text{first is}~2)P(\text{second is}~2) + P(\text{first is}~3)P(\text{second is}~1)$, keeping in mind that we already know from the problem that first number had to be either 1, 2, or 3, which gives $\frac{1}{3}\cdot \frac{1}{6} + \frac{1}{3}\cdot \frac{1}{6} + \frac{1}{3}\cdot \frac{1}{6} = \frac{1}{6}$ $\endgroup$ – JMoravitz May 24 '15 at 0:43
3
$\begingroup$

In terms of conditional probabilities, $$p(4|\text{twice})=\frac {p(4\cap \text{twice})}{p(\text{twice})}$$

$$=\frac{3\times(\frac {1}{36})}{\frac 12}=\frac 16$$

$\endgroup$
2
$\begingroup$

Note that if $X=4$ is the event that the final score is four, $N=1$ is the event that the die is rolled once, and $N=2$ is the event that the die is rolled twice, then

$$P(X=4) = P(X=4\mid N=1)P(N=1) + P(X=4\mid N=2)P(N=2).$$

Solving for $P(X=4\mid N=2)$,

$$P(X=4\mid N=2) = \frac{P(X=4) - P(X=4\mid N=1)P(N=1)}{P(N=2)}.$$

You calculated that $P(X=4) = \frac14,$ and that $P(X=4\mid N=1) = \frac13.$ We can easily find that $P(N=1) = \frac12 = P(N=2),$ so

$$ P(X=4\mid N=2) = \frac{\frac14 - \frac13 \cdot \frac12}{\frac12} = \frac16.$$

That is, if you can solve any two of the three parts of the problem correctly, then you can use the law of total probability to solve for the third part.

$\endgroup$
1
$\begingroup$

I think it is clearer to break up 1st and 2nd rolls into separate entities:

$i)1st:$ sample space:${6\choose 1}=6$. Probability that is $4$:$\space {|(4)|\over 6}={1\over 6}.\space$$ 2nd:\space$sample space:$\space$all possible $1st$ rolls combined with all possible $2nd$ rolls:$\space {6\choose 1} {6\choose 1}=6*6=36.$ Probability that adds up to $4$:$\space {|(1,3),(2,2),(3,1)|}\over 36$$={3\over 36}={1\over 12};\space$${2\over 12}+{1\over 12}={3\over 12}={1\over 4}.\space$ Are separate/disjoint therefore 'or'/added.

$ii)1st:$ sample space:${3\choose 1}=3;\space$Probability that is $4$:$\space{|(4)|\over 3}={1\over 3};\space 2nd:\space$n/a

$iii)1st:$ n/a;$\space\space 2nd:\space$sample space:$\space 1st$ roll must be $\in \{1,2,3\};\space$${3\choose 1}{6\choose 1}=3*6=18;$$\space{|(1,3),(2,2),(3,1)|}\over 18$$={3\over 18}={1\over 6}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.