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The synthetic division is not same as the polynomial long division.
Given:
Let's take a polynomial function f(x) = divided by x-2.
Using synthetic divison:
2 1 2 -5 -6
0 2 8 6
1 4 3 0
(x-2)(x^2+4x+3)
= (x-2)(x+3)(x+1)
Polynomial long division:
x-2) (x^2+4x+3
x^3- 2x^2
4x^2 - 5x + 6
4x^2 - 8x
3x + 6
3x - 6
0
=(x-2)(x^2+4x+3)
Learn more about the division here:
#SPJ4
Check the tree diagram of the problem:
The first ball can be either Red or Blue, the second ball can either be Red or Blue.
The branches represent all possible scenarios.
For example, consider the upper branch, the case when both balls are blue:
The first ball is Blue, probability 5/10, the second ball is Blue, probability 4/9 because now there are 4 left blue balls, and 9 balls in total, after we picked the first blue.
The other branches have the same logic.
The question is not completed so we have 2 cases:
"Find the probability that at least 1 ball was red, given that the first ball was red"
here we add the separate probabilities of (R, R) and (R, B)
P(R,R)+P(R,B)=2/9 + 5/18 = 4/18 +5/18= 9/18=0.5
"Find the probability that at least 1 ball was red, given that the first ball was blue"
probability is P(B, R)=5/18=0.28
The first ball can be either Red or Blue, the second ball can either be Red or Blue.
The branches represent all possible scenarios.
For example, consider the upper branch, the case when both balls are blue:
The first ball is Blue, probability 5/10, the second ball is Blue, probability 4/9 because now there are 4 left blue balls, and 9 balls in total, after we picked the first blue.
The other branches have the same logic.
The question is not completed so we have 2 cases:
"Find the probability that at least 1 ball was red, given that the first ball was red"
here we add the separate probabilities of (R, R) and (R, B)
P(R,R)+P(R,B)=2/9 + 5/18 = 4/18 +5/18= 9/18=0.5
"Find the probability that at least 1 ball was red, given that the first ball was blue"
probability is P(B, R)=5/18=0.28
Answer:
Step-by-step explanation: