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soldier1979 [14.2K]
3 years ago
14

You roll two standard number cubes. What is the probability that the sum is odd, given than one of the number cubes shows a 1?

Mathematics
1 answer:
Virty [35]3 years ago
6 0
 It's not specified whether 1 is the 1st or 2nd roll: HOWER:

The 1st Roll is "1": P(odd sum/the 1st Roll is 1)
What is the sample space of all numbers starting with "1":
{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),} = 6
the couple of add sum=(1,2), (1,4), (1,6), =3
P(odd sum/ 1st is 1) = 3/6 =1/2
or in applying the formula:

P(odd sum/the 1st Roll is 1) =P(odd sum ∩ 1) / P(getting "1") it will give the same probability = 1/2

NOW if the 2nd Roll is "1", it 's still 1/2

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Answer:

A = (32.7,47.7)

Step-by-step explanation:

Given

M = (25.5,60.1)

C = (18.3,72.5)

Required

Determine the coordinates of A

Midpoint is calculated as thus:

M(x,y) = (\frac{A_x + C_x}{2}, \frac{A_y + C_y}{2})

Substitute the given values

(25.5,60.1) = (\frac{A_x + 18.3}{2}, \frac{A_y + 72.5}{2})

Multiply through by 2

2 * (25.5,60.1) = (\frac{A_x + 18.3}{2}, \frac{A_y + 72.5}{2}) * 2

2 * (25.5,60.1) = (A_x + 18.3, A_y + 72.5)

(51,120.2) = (A_x + 18.3, A_y + 72.5)

By comparison:

A_x + 18.3 = 51

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A_x = 51 - 18.3

A_x = 32.7

A_y + 72.5 = 120.2

A_y = 120.2 - 72.5

A_y = 47.7

Hence:

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