suppose that we are going to roll two fair 6- sided dice. Find the probability that both dice show a 3. (look at the information provided within the picture.) chose 1 answer choice: A.) P(both 3)= 1\2 B.) P(both 3)= 1\3 C.) P(both 3)= 1\12 D.) P(both 3)= 1\36
suppose that we are going to roll two fair 6- sided dice. Find the probability that both dice show a 3. (look at the information provided within the picture.) chose 1 answer choice: A.) P(both 3)= 1\2 B.) P(both 3)= 1\3 C.) P(both 3)= 1\12 D.) P(both 3)= 1\36
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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I had to correct this question so please answer.
Please Answer correctly and read carefully deals with the general multiplication rule in
question:
suppose that we are going to roll two fair 6- sided dice. Find the probability that both dice show a 3. (look at the information provided within the picture.)
chose 1 answer choice:
A.) P(both 3)= 1\2
B.) P(both 3)= 1\3
C.) P(both 3)= 1\12
D.) P(both 3)= 1\36

Transcribed Image Text:Independent events: Flipping a coin twice
What is the probability of flipping a fair coin and getting "heads" twice in
a row? That is, what is the probability of getting heads on the fırst flip
AND heads on the second flip?
Imagine we had 100 people simulate this and flip a coin twice. On
average, 50 people would get heads on the first flip, and then 25 of them
would get heads again. So 25 out of the original 100 people – or 1/4 of
them – would get heads twice in a row.
The number of people we start with doesn't really matter. Theoretically,
1/2 of the original group will get heads, and 1/2 of that group will get
heads again. To find a fraction of a fraction, we multiply.
We can represent this concept with a tree diagram like the one shown
below.
1
1
1
1/2
H
%3D
4
H
1/2
1
1
1
1/2
T
%3D
2
4
1
1
1
1/2
H
4
1/2
1
1
1
1/2
T
2 2
4

Transcribed Image Text:We multiply the probabilities along the branches to find the overall
probability of one event AND the next even occurring.
For example, the probability of getting two "tails" in a row would be:
1
P(T and T)
1
2 2
4
When two events are independent, we can say that
P(A and B) = P(A) · P(B)
Be careful! This formula only applies to independent events.
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