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Vikentia [17]
3 years ago
12

A television game has 6 shows doors, of which the contests must pick 2. behind two of the doors are expensive cars, and behind t

he other 4 doors are consolation prizes. Find the probability that the contestant wins exactly 1 car? no car? or atleast one car?
what if they had to pick 3 doors? and all of the other information wast the same
Mathematics
1 answer:
Katyanochek1 [597]3 years ago
3 0
The answer to this question:
One car probability 82/120
No car probability = 24/120
At least one car probability= 96/120

I will focus answering the 3 doors probability since the 2nd door problem is solved in the previous problem. (brainly.com/question/5761449)

No car condition
1. 1st door consolation, 2nd door consolation=, 3rd door consolation= 4/6 * 3/5 * 2/4= 24/120
This was also can be found by: (4!/1!)/ (6!/3!) = 24/120

(At least one car probability)  is the opposite of (no car probability) In this case, the easier way is 
100% - (no car probability) = 120/120 - 24/120= 96/120

One car probability is (At least one car probability) - (2 car probability). It will be easier to count the 2 car probability and subtract the (At least one car probability) 
Two car condition:
1. 1st door car, 2nd door car, 3rd door consolation = 2/6 * 1/5 * 4/4 =8/ 120
2.1st door car, 2nd door consolation, 3rd door car =2/6 * 4/5 * 1/4 = 8/120
3. 1st door consolation, 2nd door car, 3rd door car= 4/6 * 2/5 * 1/4= 8/120
The total probability will be 8/120+ 8/120 + /120= 24/120
This was also can be found by: (2!) (4!/2!)/ (6!/3!) = 24/120

One car probability =  (At least one car probability) - (2 car probability)= 96/120-24/120= 82/120
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A point $(x, y)$ with integer coordinates is randomly selected such that $0 \le x \le 8$ and $0 \le y \le 4$. what is the probab
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Answer:

\frac{1}{3}

Step-by-step explanation:

A point (x, y) with integer coordinates is randomly selected such that 0 \le x \le 8 \:and\: $0 \le y \le 4$.

The possible pairs of (x,y) are:

(0,0),(0,1),(0,2),(0,3),(0,4)

(1,0),(1,1),(1,2),(1,3),(1,4)

(2,0),(2,1),(2,2),(2,3),(2,4)

(3,0),(3,1),(3,2),(3,3),(3,4)

(4,0),(4,1),(4,2),(4,3),(4,4)

(5,0),(5,1),(5,2),(5,3),(5,4)

(6,0),(6,1),(6,2),(6,3),(6,4)

(7,0),(7,1),(7,2),(7,3),(7,4)

(8,0),(8,1),(8,2),(8,3),(8,4)

The Total Possible Outcomes n(S)= 45

The pair (x, y) that satisfies the given condition (say event A: x + y \le 4) are:

(0,0),(0,1),(0,2),(0,3),(0,4)\\(1,0),(1,1),(1,2),(1,3)\\(2,0),(2,1),(2,2)\\(3,0),(3,1)\\(4,0)

n(A)=15

Therefore:

P(A)=\frac{n(A)}{n(S)} =\frac{15}{45} =\frac{1}{3}

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The sum of two numbers is 36. The first number is one fifth 1 /5 of the second number. What are the​ numbers?
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The two numbers are 30 and 6
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The polynomial 3x2 is a with a degree of .
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To determine the degree of a polynomial, you look at every term:

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So, for example, the degree of 7x^{55} is 55, while the degree of xy^2z^4 is 1+2+4 = 7

Finally, the term of the degree of the polynomial is the highest degree among its terms.

So, 3x^2 is a degree 2 polynomial (although it only has one term)

similarly, x^2y + 3xy^2 + 1 is a degree 3 polynomial: the first two terms have degree 3, because they have exponents 2 and 1.

6 0
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Equivalent fractions to 1/3
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= 2/6 = 3/9 = 12/36 = 24/72 
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