# what is the PROBABILITY OF THIS?

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#### Dencology

##### Full Member
10+ Year Member
15+ Year Member
what is the probability of throwing two fours with a dice in five tries?

ans.
10(125)/36(216)

how?

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

ans.
10(125)/36(216)

how?
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 theres 1 4 in a 6 faced 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=5 attemps
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 get
5(2)(1/6^2)(5/6^(5-2))=
(10) (1/36) (125/216)= (10)(1)(125)/ (36)(216)

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.

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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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

Hey guys, I just want to ask you, what's the probability if you use 2 dies?

They are independent events so it won't matter. The thing is you'd have to roll 2 dice two times (4 rolls total) and then take one die and roll it once for the 5th roll.

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 theres 1 4 in a 6 faced 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=5 attemps
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 get
5(2)(1/6^2)(5/6^(5-2))=
(10) (1/36) (125/216)= (10)(1)(125)/ (36)(216)

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)

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

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)

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.

Heres a question for streetwolf that he will find quite challenging.

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 3*.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?

Heres a question for streetwolf that he will find quite challenging.

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.
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).

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.

how did you get 3*.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?
to be honest, i dont get it either, why not 3 & not 6 * .49 * .3??