Answer to Question #210113 in Electrical Engineering for Bahawal Tahir

Question #210113

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?


1
Expert's answer
2021-06-24T15:40:02-0400

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}"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS