what is the probability of throwing two fours with a dice in five tries?

ans.

10(125)/36(216)

how?

ans.

10(125)/36(216)

how?

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- Thread starter Dencology
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what is the probability of throwing two fours with a dice in five tries?

ans.

10(125)/36(216)

how?

ans.

10(125)/36(216)

how?

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i HATE these problems!!!what is the probability of throwing two fours with a dice in five tries?

ans.

10(125)/36(216)

how?

So they use this stupid formula where you have to figure out the probability of getting a 4 when you throw a dice. It is 1/6, right? Bc theres

Now the probability of NOT getting a 4 in 1 throw is (5/6) (five other numbers- 1,2,3,5,6 in a die)

nr(p^r)(q^(n-r))

n=

r= success out of 5 attempts=

p=prob of success in 1 throw=

q= prob. of failure in 1 throw= 1-p= 1-(1/6) =

Since we have to throw a four 2 times, we bring (

Since we have failure the rest of the times (5times total- 2successful times= 3 times of failure) we bring up the failure to the power of 3

So, we get

(10) (1/36) (125/216)= (10)(1)(125)/ (36)(216)

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what is the probability of throwing two fours with a dice in five tries?

ans.

10(125)/36(216)

how?

Choose the 2 rolls that are 4s = (5 C 2), account for the 2 successes (1/6)^2 and account for the 3 failures (5/6)^3.

(5 C 2) = 10.

You get 10(125)/(36)(216)

The 10 is from (5 C 2), the 125 is 5^3, the 36 is 6^2, and the 216 is 6^3.

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Choose the 2 rolls that are 4s = (5 C 2), account for the 2 successes (1/6)^2 and account for the 3 failures (5/6)^3.

(5 C 2) = 10.

You get 10(125)/(36)(216)

The 10 is from (5 C 2), the 125 is 5^3, the 36 is 6^2, and the 216 is 6^3.

Amazing, abolutely amazing. I swear I have never seen a math question that Streetwolf did NOT answer in the year and a half I have been on SDN.

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This thread has another of the same type of question.

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Amazing, abolutely amazing. I swear I have never seen a math question that Streetwolf did NOT answer in the year and a half I have been on SDN.

There have definitely been a few. Just a few

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i HATE these problems!!!

So they use this stupid formula where you have to figure out the probability of getting a 4 when you throw a dice. It is 1/6, right? Bc theres14 in a6faced die.

Now the probability of NOT getting a 4 in 1 throw is (5/6) (five other numbers- 1,2,3,5,6 in a die)

So we have a formula Prob of r success out of n attempts=

nr(p^r)(q^(n-r))

n=5attemps

r= success out of 5 attempts=2

p=prob of success in 1 throw=1/6

q= prob. of failure in 1 throw= 1-p= 1-(1/6) =5/6

Since we have to throw a four 2 times, we bring (1/6)^2

Since we have failure the rest of the times (5times total- 2successful times= 3 times of failure) we bring up the failure to the power of 3(5/6)^3

So, we get5(2)(1/6^2)(5/6^(5-2))=

(10) (1/36) (125/216)= (10)(1)(125)/ (36)(216)

so i did what you said:

3(2)(7/10)^2(3/10)^1

which it comes down to

6(49/100)(3/10)

but the answer is

3(49/100)(3/10). how?

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If there is a 70% chance of rain each day for the next 3 days. what is the probability it will rain 2 out of the 3 days?

so i did what you said:

3(2)(7/10)^2(3/10)^1

which it comes down to

6(49/100)(3/10)

but the answer is

3(49/100)(3/10). how?

3c2 * .7^2 * .3^1

This equals 3*.49 * .3 which is the correct answer. You calculated 3c2 wrong.

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if you have a branch that is 10 inches in diameter and there is a rope that spirals around it. The vertical distance between each complete revolution is 35 inches. How much rope is required to make on complete revolution.

(remember that the rope is spiralling up the branch so you need to account for the thickness of the branch as well is the vertical height of the revolution). I think you may enjoy this one streetwolf.

3c2 * .7^2 * .3^1

This equals 3*.49 * .3 which is the correct answer. You calculated 3c2 wrong.

how did you get

r= success out of 3 attempts=

p=prob of success in 1 throw=

q= prob. of failure in 1 throw= 1-p= 1-(0.7) =

i followed this and still the answer would come out wrong. Why?

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If you unravel the outer coating of the branch with the rope still attached, within the area of exactly 1 revolution, what you'd get is a rectangular sheet of the branch with height (vertical) = 35 inches and length (horizontal) = 10pi inches (this is the circumference of the branch). The rope itself would run on the diagonal from bottom left to top right (or bottom right to top left, whatever you choose).

if you have a branch that is 10 inches in diameter and there is a rope that spirals around it. The vertical distance between each complete revolution is 35 inches. How much rope is required to make on complete revolution.

(remember that the rope is spiralling up the branch so you need to account for the thickness of the branch as well is the vertical height of the revolution). I think you may enjoy this one streetwolf.

Either way, the length of the rope is found with a^2 + b^2 = c^2. Here we have a = the height and b = the width. You get c = 47.03 inches.

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to be honest, i dont get it either, why nothow did you get3*.49 * .3. i don't get this three. i thought i would be 6 because of this formula:nr(p^r)(q^(n-r))

n=3 attemps

r= success out of 3 attempts=2

p=prob of success in 1 throw=70%

q= prob. of failure in 1 throw= 1-p= 1-(0.7) =0.3

i followed this and still the answer would come out wrong. Why?

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