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
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 ways to select 3 sides from the 4,
then P(A)=
For B
the 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
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
Hello, thereddevils!
We don't need factorials for this problem.
On one roll, we have these outcomes: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 .
. . : the score T at the first toss is 7
. . : the sum of first 2 scores is less than 15
. . : the product of the three scores is a multiple of 3.
FInd: .
. .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
With 2 rolls, there are: . outcomes.
If the sum is less than 15, we have:
. .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
With three rolls, there are: . 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.
. .
Hence: .
Therefore: .