A six-sided die is rolled five times. What is the probability that only the final roll will be a deuce?
I've tried to reason this out myself but I can only think that there's a 1/6 chance that the roll will be 2 and another 1/5 chance for it to be the last one. What am I missing here?
Thanks!
$\endgroup$3 Answers
$\begingroup$Allow me to quote your reasoning:
1/5 chance for it to be the last one
Here is what is wrong with your reasoning: it is not guaranteed that only one roll will be a deuce. It is also possible that there are two deuces rolled.
We need the first four rolls to be not deuce and the last one to be a deuce. Each roll is independent, so we can multiply the probabilities for each roll together.
The probability for a roll to be a deuce is $\dfrac16$, and similarly the probability for a roll to be not a deuce is $\dfrac56$. Therefore, the required probability is $\left(\dfrac56\right)^4\left(\dfrac16\right) = \dfrac{625}{7776}$.
$\endgroup$ $\begingroup$The first $4$ rolls are not $2$ and the $5$th rolls is $2$. So the probability is $$\left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)$$
$\endgroup$ $\begingroup$You need roll one to be one of 1, 3, 4, 5, 6. What is the probability of that?
You need roll two to be one of 1, 3, 4, 5, 6. What is the probability of that?
You need roll three to be one of 1, 3, 4, 5, 6. What is the probability of that?
You need roll four to be one of 1, 3, 4, 5, 6. What is the probability of that?
You need roll five to be 2. What is the probability of that?
What is the probability of all these events occurring?
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