Question

Q7

Suppose that two balanced dice are tossed repeatedly and the sum of the twouppermost faces is determined on cach toss. What is the probability that we obtana

.a) sum of 3 before we obtaim a sum of 7?

b). sum of 4 before we obtain a sum of 7?

Answer 1

Part a

First of all, we define the events as:

A: a sum of 3

B: not a sum of 3 or 7 

Since there are 36 possible rolls, $P(A) = \frac{2}{36}$ and $P(B) = \frac{28}{38}$ . Obtaining a sum of 3 before a sum of 7 can happen on the first roll, the 2nd roll, the 3rd roll, etc.

$\frac{2}{36}+ \frac{28}{36}* \frac{2}{36}+ (\frac{28}{36})^2* \frac{2}{36}+.........$

$=\frac{2}{36}*\frac{1}{(1-\frac{28}{36})}=\frac{1}{4}$

Part b

First of all, we define the events as:

A: a sum of 4

B: not a sum of 4 or 7 

Since there are 36 possible rolls, $P(A) = \frac{3}{36}$ and $P(B) = \frac{27}{38}$ . Obtaining a sum of 4 before a sum of 7 can happen on the first roll, the 2nd roll, the 3rd roll, etc.

$\frac{3}{36}+ \frac{27}{36}* \frac{3}{36}+ (\frac{27}{36})^2* \frac{3}{36}+................$

$=\frac{3}{36}*\frac{1}{(1-\frac{27}{36})}=\frac{1}{3}$