Question

We roll a fair four-sided die. If the result is $1$ or $2$, we roll once more but otherwise, we stop. What is the probability that the sum total of our rolls is at least $4$?

• A. $\frac{1}{16}$
• B. $\frac{4}{16}$
• C. $\frac{7}{16}$
• D. $\frac{9}{16}$

Answer 1

The correct option is D $\frac{9}{16}$

$A_i$: the result of the first roll is 𝑖

$P\left(A_i\right)=\frac{1}{4},i=1,2,3,4.$

𝐵: the sum total is at least 4.

$P\left(B\right)=su{m}_{i=1}^{4}P\left(A_i\right)P\left(B/A_i\right)$

Given:

$A_1$: the sum total will be ≥4 if the second roll results in 3 or 4, which happens with probability $\frac{1}{2}$.

Thus $P\left(B/A_1\right)=\frac{1}{2}$,

Similarly $P\left(B/A_2\right)=\frac{3}{4}$.

Given $A_2$, the sum total will be ≥4 if the second roll results in 2, 3, or 4, which happens with probability $\frac{3}{4}$

Given $A_3$: you stop and the sum total remains below 4.

Thus $P\left(B/A_3\right)=0,$

Given $A_4$: you stop but the sum total is already 4.

Thus $P\left(B/A_4\right)=1$.

By the total probability theorem

$P\left(B\right)=\frac{1}{4}\times \frac{1}{2}+\frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times 0+\frac{1}{4}\times 1=\frac{9}{16}$