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IceJOKER [234]
3 years ago
13

You roll 4 fair, distinguishable, six-sided dice. What is the probability of rolling :______

Mathematics
1 answer:
Vikentia [17]3 years ago
6 0

Answer:

a)0.25

b)0.375

Step-by-step explanation:

Possible outcomes when dice are rolled : 1 ,2 ,3,4,5 ,6

Even= 2,4,6  

Odd = 1 ,3 , 5

Probability of getting even =\frac{3}{6}=\frac{1}{2}

Probability of getting odd = \frac{3}{6}=\frac{1}{2}

4 dices are rolled together

a)P(3 even and 1 odd numbers)

P(3 even and 1 odd numbers)=^4C_3 (\frac{1}{2})^3 (\frac{1}{2})^1

P(3 even and 1 odd numbers)=\frac{4!}{3!1!} (\frac{1}{2})^3 (\frac{1}{2})^1 =0.25

b)P(2 even and 2 odd numbers)

P(2 even and 2 odd numbers)=^4C_2 (\frac{1}{2})^2 (\frac{1}{2})^2

P(3 even and 1 odd numbers)=\frac{4!}{2!2!} (\frac{1}{2})^2 (\frac{1}{2})^2=0.375

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sweet [91]
To model and solve our situation we are going to use the equation: s= \frac{d}{t}
where
s is speed
d is distance 
t is time 

1. We know that the distance between the cities is 2400 miles, so d=2400. We also know that the speed of the plane is 450 mi/h. Since we don't know the speed of the air, S_{a}=?. We don't know how much the westward trip takes, so t_{w}=?, and we also don't know how much the eastward trip takes, so t_{e}=?.

Going westward. Here the plane is flying against the air, so we need to subtract the speed of the air from the speed of the plane:
450-S_{a}= \frac{2400}{t_{w} }
Going eastward. Here the plane is flying with the the air, so we need to add the speed of the air to the speed of the plane:
450+S_{a}= \frac{2400}{t_{e} }

2. We know for our problem that the round trip takes 11 hours; so the total time of the trip is 11, t_{t}=11. Notice that we also know that the total time of the trip equals time of the tip going westward plus time of the trip going eastward, so t_{t}=t_{w}+t_{e}. Since we know that the total trip takes 11 hours, we can replace that value in our total time equation and solve for t_{w}:
11=t_{w}+t_{e}
t_{w}=11-t_{e}

Now we can replace t_{w} in our going westward equation to model our round trip with a system of equations:
450-S_{a}= \frac{2400}{t_{w}}
450-S_{a}= \frac{2400}{11-t_{e} } equation (1)
450+S_{a}= \frac{2400}{t_{e}} equation (2)

3. To solve our system of equations, we are going to solve for t_{e} in equations (1) (2):

From equation (1)
450-S_{a}= \frac{2400}{11-t_{e} }
11-t_{e}= \frac{2400}{450-S_{a} }
-t_{e}= \frac{2400}{450-S_{a} } -11
t_{e}=11- \frac{2400}{450-S_{a} }
t_{e}= \frac{4950-11S_{a} -2400}{450-S_{a} }
t_{e}= \frac{2550-11S_{a} }{450-S_{a} } equation (3)

From equation (2):
450+S_{a}= \frac{2400}{t_{e} }
t_{e}= \frac{2400}{450+S_{a} } equation (4)

Replacing (4) in (3)
\frac{2400}{450+S_{a}} = \frac{2550-11S_{a}}{450-S_{a} }
Now, we can solve for S_{a} to find the speed of the wind:
2400(450-S_{a})=(450+S_{a})(2550-11S_{a})
1080000-2400S_{a}=1147500-4950S_{a}+2550S_{a}-11(S_{a})^{2}
11(S_{a})^{2}-67500=0
11(S_{a})^{2}=67500
(S_{a})^{2}= \frac{67500}{11}
S_{a}=+/-  \sqrt{ \frac{67500}{11} }
Since speed cannot be negative, the solution of our equation is:
S_{a}= \sqrt{ \frac{67500}{11} }
S_{a}=78.33

We can conclude that the speed of the wind is 78 mph.

3 0
4 years ago
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