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coldgirl [10]
4 years ago
12

An experiment consist of rolling 3 fair dice (1 red die, a blue die, and a white die) and recording the number rolled on each di

e. Assume that the dice are fair, so that all outcomes are equally likely.
1. What probability should be assigned to each outcome?
2. What is the probability that the sum of the numbers rolled is 6?
3. What is the probability that the sum of the numbers rolled is at most 5?
Mathematics
1 answer:
Setler79 [48]4 years ago
3 0
In 1, t<span>here are 6 outcomes for each die, so for three dice, the total combination is 6 x 6 x 6 = 216 outcomes. Hence, t</span><span>he probability of any individual outcome is 1/216 </span>

The outcomes that will add up to 6 are 
<span>1+1+4 </span>
<span>1+4+1 </span>
<span>4+1+1 </span>
<span>1+2+3 </span>
<span>1+3+2 </span>
<span>2+1+3 </span>
<span>2+3+1 </span>
<span>3+1+2 </span>
<span>3+2+1 </span>
<span>2+2+2 </span>

<span>Hence the probability is </span><span>10/216 </span>

In 3, the minimum sum of the three dice is 3. so we start with this 
<span>P(n = 3) </span>
<span>1+1+1 ; </span><span>1/216 </span>

<span>P(n = 4) </span>
<span>1+1+2 </span>
<span>1+2+1 </span>
<span>2+1+1 ; </span><span>3/216 </span>

<span>P(n = 5) </span>
<span>1+1+3 </span>
<span>1+3+1 </span>
<span>3+1+1 </span>
<span>1+2+2 </span>
<span>2+1+2 </span>
<span>2+2+1; </span><span>6/216 

The sum in 3 is 10/216 or 5/108</span>
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Sophie [7]
Consider the closed region V bounded simultaneously by the paraboloid and plane, jointly denoted S. By the divergence theorem,

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV

And since we have

\nabla\cdot\mathbf f(x,y,z)=1

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV
=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
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Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by D, we have

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS

Parameterize D by

\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k
\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k

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\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du
=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du
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\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2
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3 years ago
Slips of paper numbered 1 to 15 are placed in a box. A slip of paper is drawn at random. What is the probability that the number
monitta

Given:

Slips of paper numbered 1 to 15 are placed in a box.

To find:

The probability that the number picked is either a multiple of 5 or an odd number.

Solution:

We have,

Total outcomes = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

No. of total outcomes = 15

Multiple of 5 are 5, 10, 15.

Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15.

Number that are either a multiple of 5 or an odd number are 1, 3, 5, 7, 9, 10, 11, 13, 15.

No. of favorable outcomes = 9

We know that,

\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}

\text{Probability}=\dfrac{9}{15}

\text{Probability}=\dfrac{3}{5}

\text{Probability}=0.6

Therefore, the  probability that the number picked is either a multiple of 5 or an odd number is 0.6.

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#a

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The carnival always costs the least.

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#c

The aquarium always costs the move.

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