# tetrahedron die probability

• Jan 17th 2010, 01:45 AM
thereddevils
tetrahedron die probability
A fair tetrahedron die has four faces , each with digit 1,2,3,4 . THe die is tossed three times. The sum of three visible surfaces is recorded as T at each toss . The events A , B and C are defined as follows .

A= The score T at the first toss is 7 .
B=SUm of first 2 scores is less than 15
C=Product of the three scores is a multiple of 3 .

FInd P(A) , P(B) and P(C)

For P(A) , It can only (1,3,4) and this can be arranged in 3!=6 ways so
P(A)=6/24=1/4

For P(B) , its either (6,8) , (8,6) or (6,6)

6=(1,2,3) , 8=(1,3,4)

so p(B)=3 x (3!/4!)(3!/4!)=3/16 but the answer given is 3/8
• Jan 17th 2010, 04:06 AM
awkward
Quote:

Originally Posted by thereddevils
A fair tetrahedron die has four faces , each with digit 1,2,3,4 . THe die is tossed three times. The sum of three visible surfaces is recorded as T at each toss . The events A , B and C are defined as follows .

A= The score T at the first toss is 7 .
B=SUm of first 2 scores is less than 15
C=Product of the three scores is a multiple of 3 .

FInd P(A) , P(B) and P(C)

For P(A) , It can only (1,3,4) and this can be arranged in 3!=6 ways so
P(A)=6/24=1/4

For P(B) , its either (6,8) , (8,6) or (6,6)

6=(1,2,3) , 8=(1,3,4)

so p(B)=3 x (3!/4!)(3!/4!)=3/16 but the answer given is 3/8

Don't overlook rolling two 7's (7=1+2+4).
• Jan 17th 2010, 05:12 AM
For A

There is only 1 way to score a 7 on the first throw,
which is 4, 2 and 1 if all are visible.

As there are $\binom{4}{3}$ ways to select 3 sides from the 4,

then P(A)= $\frac{1}{\binom{4}{3}}
$

For B

the $\binom{4}{3}$ possibilities are 4+3+2=9, 4+3+1=8, 4+2+1=7, 3+2+1=6

Since 9+6=15, we cannot have 9.
Since 8+7=15, only 8+6 and 6+8 contain 8, but 7 can pair with 6 or itself,
giving 7+7, 7+6 and 6+7.
The last one is 6+6.

That's 6 options, so the probability is $\frac{6}{\binom{4}{3}^2}$

For C

the product of the 3 scores will be a multiple of 3 if 6 or 9 came up.
6x and 9x are multiples of 3, for x natural.

Hence, the product will not be a multiple of 3 if only 8 and 7 come up.
888, 777 are two.
All 3 arrangements of two 8's and a 7 and all 3 arrangements of two 7's and an 8 complete the list, so the probability is

$1-\frac{8}{4^3}$
• Jan 17th 2010, 05:24 AM
Soroban
Hello, thereddevils!

We don't need factorials for this problem.

Quote:

A fair tetrahedron die has four faces, each with digit 1,2,3,4.
THe die is tossed three times.
The sum of three visible surfaces is recorded as T at each toss.

The events A , B and C are defined as follows .
. . $A$: the score T at the first toss is 7
. . $B$: the sum of first 2 scores is less than 15
. . $C$: the product of the three scores is a multiple of 3.

FInd: . $P(A),\; P(B)\text{, and }P(C)$

On one roll, we have these outcomes:

. . $\begin{array}{ c c c }
\text{Visible}& \text{Sum}& \\
\text{faces} & (T) & \text{Prob.} \\ \hline \\[-3mm]
1,2,3 & 6 & \dfrac{1}{4} \\ \\[-3mm]
1,2,4 & 7 & \dfrac{1}{4} \\ \\[-3mm]
1,3,4 & 8 & \dfrac{1}{4} \\ \\[-3mm]
2,3,4 & 9 & \dfrac{1}{4} \\ \\[-3mm] \hline\end{array}$

$(a)\;P(A) \;=\;P(\text{sum is 7}) \;=\;\frac{1}{4}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

With 2 rolls, there are: . $4^2 = 16$ outcomes.

If the sum is less than 15, we have:

. . $\begin{array}{ccc}
\text{Rolls} & \text{Sum} & \text{Prob.} \\ \hline \\[-3mm]
(6,6) & 12 & \dfrac{1}{16} \\ \\[-3mm]
(6,7), (7,6) & 13 & \dfrac{2}{16} \\ \\[-3mm]
(6,8),(7,7),(8,6) & 14 & \dfrac{3}{16} \\ \\[-3mm]\hline \end{array}$

$(b)\;P(B) \;=\; P(\text{sum is less than 15}) \;=\;\frac{1}{16} + \frac{2}{16} + \frac{3}{16} \;=\;\frac{6}{16} \;=\;\frac{3}{8}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

With three rolls, there are: . $4^3 = 64$ outcomes.

If the product of three rolls is a multiple of 3,
. . at least one of the rolls is a multiple of 3.

How many are not a multiple of 3?
. . They must be comprised of 7's and/or 8's.

. . $\begin{array}{cc}
\text{Rolls} & \text{Prob.} \\ \hline \\[-3mm]
(7,7,7) & \dfrac{1}{64} \\ \\[-3mm]
(7,7,8) & \dfrac{3}{64} \\ \\[-3mm]
(7,8,8) & \dfrac{3}{64} \\ \\[-3mm]
(8,8,8) & \dfrac{1}{64} \\ \\[-3mm] \hline \end{array}$

Hence: . $P(not\text{ multiple of 3}) \;=\;\frac{1}{64} + \frac{3}{64} + \frac{3}{64} + \frac{1}{64} \;=\;\frac{8}{64} \;=\;\frac{1}{8}$

Therefore: . $P(\text{multiple of 3}) \;=\;1-\frac{1}{8} \;=\;\frac{7}{8}$