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