Several probability problems and their solutions

Jan 2020
11
0
Bulgaria
Problem 1


Three 6 sided dice are thrown. What's the probability that the sum of points is 11? And 12?

The given solution is P{11} = 27/216, P{12} = 25/216

My solution is

P{11} = (1,5,5),(5,5,1),(2,4,5),(5,4,2)/6^3 = (1,5,5),(5,5,1),(2,4,5),(5,4,2)/216

P{12} = (1,5,6),(6,5,1),(2,4,6),(6,4,2),(3,4,5),(5,4,3)/6^3 = (1,5,6),(6,5,1),(2,4,6),(6,4,2),(3,4,5),(5,4,3)/216


Problem 2

n six sided dice are thrown. Calculate the probability, that the sum of points that come up is no lesser than 6n - 1

The given solution is n+1/6^n
I don't understand this one

Problem 3

Two six sided dice are thrown. What's the probability that the sum of points is an even number?

The given solution is 3x3+6x3/6^2 = 0,75

My solution is based on a previous thread, with the difference that one die is thrown 2 times
2,4,6,8,10,12
There are:
(1,1) ways to get 2
(1,3)(3,1)(2,2) to get 4
(1,5)(5,1)(2,4)(4,2)(3,3) to get 6
(3,5)(5,3)(2,6)(6,2)(4,4) to get 8
(4,6)(6,4)(5,5) to get 10
(6,6) to get 12
=> 1+3+5+5+3+1/36 = 18/36 = 1/2 = 0,5

Problem 4

From an urn, which contains balls with the numbers 1,2,...,N, we subtract 1 ball n times. Calculate the probability that the numbers of the balls that are subtracted, written in the order of subtraction, form a growing row, if after every subtraction, the subtracted ball:
a) returns to the urn before the next subtraction
b) doesn't return to the urn

The given solution is:
a) CN^k/N^k
b) 1/n!

Don't understand this one either

Problem 5

n dice are thrown. Calculate the probability that n1 ones, n2 twos,...,n6 sixes come up, when n1+n2+...+n6 = n

The given solution is
n!/n1!*n2!*...*n6!*6^n

I don't understand this one
 

romsek

Math Team
Sep 2015
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1,673
USA
Problem 1


Three 6 sided dice are thrown. What's the probability that the sum of points is 11? And 12?

The given solution is P{11} = 27/216, P{12} = 25/216

My solution is

P{11} = (1,5,5),(5,5,1),(2,4,5),(5,4,2)/6^3 = (1,5,5),(5,5,1),(2,4,5),(5,4,2)/216

P{12} = (1,5,6),(6,5,1),(2,4,6),(6,4,2),(3,4,5),(5,4,3)/6^3 = (1,5,6),(6,5,1),(2,4,6),(6,4,2),(3,4,5),(5,4,3)/216
The answer you gave, listing the rolls out then dividing by 216, wouldn't get you any credit. You have to count the number of rolls, not just list them.

Additionally if you are going to count every possible 3 dice roll, as you do by having 216 as the denominator, then you must count all the permutations of the desired rolls.

For example you correctly list (1,5,6) as a roll that sums to 12. Well so do (5,6,1), (6,1,5), (1,6,5), (6,5,1), and (5,1,6).

Try this again and see if you can come up with the given answers.
 
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romsek

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Sep 2015
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="whois1230, post: 617159,
Problem 2

n six sided dice are thrown. Calculate the probability, that the sum of points that come up is no lesser than 6n - 1

The given solution is n+1/6^n
I don't understand this one
The maximum roll sum is going to be $6n$
So we want the probability of rolling $6n-1$ or $6n$

There is only one way to roll $6n$, all sixes.
There are $n$ ways to roll $6n-1,~n-1$ sixes, and $1$ five. There are $n$ slots for the five.

$P[6n]=\dfrac{1}{6^n}$

$P[6n-1] = \dfrac{n}{6^n}$

$P[n]+P[6n-1] = \dfrac{n+1}{6^n}$
 

romsek

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Sep 2015
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USA
Problem 5

n dice are thrown. Calculate the probability that n1 ones, n2 twos,...,n6 sixes come up, when n1+n2+...+n6 = n

The given solution is
n!/n1!*n2!*...*n6!*6^n

I don't understand this one
\(\displaystyle P = \dfrac{\dbinom{n}{n1}\dbinom{n-n1}{n2}\dbinom{n-n1-n2}{n3}\dbinom{n-n1-n2-n3}{n4}\dbinom{n-n1-n2-n3-n4}{n5}\dbinom{n-n1-n2-n3-n4-n5}{n6}}{6^n}\)

See if you can understand why, and take a crack at reducing it. You'll note that there cancellation between adjacent terms.
 
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romsek

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Sep 2015
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Problem 3

Two six sided dice are thrown. What's the probability that the sum of points is an even number?

The given solution is 3x3+6x3/6^2 = 0,75

My solution is based on a previous thread, with the difference that one die is thrown 2 times
2,4,6,8,10,12
There are:
(1,1) ways to get 2
(1,3)(3,1)(2,2) to get 4
(1,5)(5,1)(2,4)(4,2)(3,3) to get 6
(3,5)(5,3)(2,6)(6,2)(4,4) to get 8
(4,6)(6,4)(5,5) to get 10
(6,6) to get 12
=> 1+3+5+5+3+1/36 = 18/36 = 1/2 = 0,5
The given answer is incorrect. It should be intuitively obvious that the answer is 1/2.
Dice have 3 even faces and 3 odd faces. The number of rolls summing even will be equal to the number of rolls summing odd.

If the first roll is even. The second must be even. If the first roll is odd, the second must be odd.

$P[E]=P[O]=\dfrac 1 2$

$P= P[EE]+P[OO] = \left(\dfrac 1 2\right)^2 + \left(\dfrac 1 2\right)^2 = 2 \cdot \dfrac 1 4 = \dfrac 1 2$
 
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romsek

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#4 is eluding me at the moment and I have to go to work.
 
Jan 2020
11
0
Bulgaria
The answer you gave, listing the rolls out then dividing by 216, wouldn't get you any credit. You have to count the number of rolls, not just list them.

Additionally if you are going to count every possible 3 dice roll, as you do by having 216 as the denominator, then you must count all the permutations of the desired rolls.

For example you correctly list (1,5,6) as a roll that sums to 12. Well so do (5,6,1), (6,1,5), (1,6,5), (6,5,1), and (5,1,6).

Try this again and see if you can come up with the given answers.
P{11} = (1,5,5),(5,1,5),(5,5,1),(1,4,6),(4,1,6),(6,1,4),(6,4,1),(1,6,4),(4,6,1),(2,3,6),(3,2,6),(6,2,3),(2,6,3),(3,6,2),(6,3,2),(3,3,5),(3,5,3),(5,3,3),
(2,4,5),(2,5,4),(4,2,5),(4,5,2),(5,2,4),(5,4,2),(3,4,4),(4,3,4),(4,4,3)/6^3 = 27/216

P{12} = (1,5,6),(1,6,5),(5,1,6),(5,6,1),(6,5,1),(6,1,5),(2,4,6),(2,6,4),(4,2,6),(4,6,2),(6,2,4),(6,4,2),(3,4,5),(3,5,4),(4,3,5),(4,5,3),(5,3,4),(5,4,3),
(5,5,2),(5,2,5),(2,5,5),(4,4,4),(3,6,6),(6,3,6),(6,6,3)/6^3 = 25/216

Isn't there a shortcut to do this, or some formula?
 

romsek

Math Team
Sep 2015
2,958
1,673
USA
Problem 4

From an urn, which contains balls with the numbers 1,2,...,N, we subtract 1 ball n times. Calculate the probability that the numbers of the balls that are subtracted, written in the order of subtraction, form a growing row, if after every subtraction, the subtracted ball:
a) returns to the urn before the next subtraction
b) doesn't return to the urn

The given solution is:
a) CN^k/N^k
b) 1/n!
The answer given in (b) is correct only if $N=n$
The answer reflects the fact that only one of the $n!$ permutations of a list of $n$ unique integers appears in order.

If we assume that for part (a) that $N=n$ as well then we have the probability of picking the first $n-1$ balls, in order, in $n-1$ picks
Thus at each pick we must pick the correct ball for that pick.

\(\displaystyle p = \left(\dfrac{1}{n}\right)^{n-1} = \dfrac{1}{n^{n-1}}\)

I can't interpret what the given solution for (a) is but I've verified the above number by sim and it is correct.
 
Jan 2020
11
0
Bulgaria
The answer given in (b) is correct only if $N=n$
The answer reflects the fact that only one of the $n!$ permutations of a list of $n$ unique integers appears in order.

If we assume that for part (a) that $N=n$ as well then we have the probability of picking the first $n-1$ balls, in order, in $n-1$ picks
Thus at each pick we must pick the correct ball for that pick.

\(\displaystyle p = \left(\dfrac{1}{n}\right)^{n-1} = \dfrac{1}{n^{n-1}}\)

I can't interpret what the given solution for (a) is but I've verified the above number by sim and it is correct.
Problem 6

10 distinguishable dice are thrown. What's the probability that an equal number of "ones" and "sixes" come up?

Problem 7

3 distinguishable even dice are thrown. What's the probability of the event A = {the sum and the product of the numbers that come up are equal}?

Problem 8

From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. What's the probability that the ball, which was taken out is:
a) white
b) green
c) red

Problem 9

An urn contains 8 white and 4 black balls. Two balls are simultaneously taken out. What is more probable: that the two balls are white or that the two
balls are with a different color?

Problem 10

From an urn, which contains 12 white and 8 black balls, 2 balls are simultaneously taken out. Find the probability of the events:
A={both balls are white}
B={both balls are black}
c={the two balls have different colors}

Problem 11

From an urn, which contains M white and N black balls, 2 balls are simultaneously taken out. Find the probability of the events:
A={both balls are white}
B={both balls are black}
C={the two balls have different colors}

Problem 12

In an urn there are 2*M white and 2*N black balls. M+N balls are simultaneously subtracted. What's the probability that M white and N black balls remain
in the urn?
 

romsek

Math Team
Sep 2015
2,958
1,673
USA
If you're going to add more questions please do it by posting a new thread.

So these are 6 new problems? Once again I ask you to try them first and report back with where you get stuck.