1. ## Probability Questions

Hello Guys ,
I am facing problem to solve Q 9,10. Please if any one could help me understand these problems and confirm my answer of Q 8 or correct me if I am wrong. I will be thankful

8. A and B each throw a set of three dice. If A throws to get a total of 16, B’s chance of getting a higher total is

(A) 1/54
(B) 1/36
(C) 5/108
(D) 53/54
since ,we have 3 dices for B and we need to get sum >16
so the options are 6*6*6, 6*6*5,
and all possibilities of an o/p of 3 dices is 3*(6+6+6) =54
then All Possibilities of A+ B= 108
B chance of getting higher total is =2/108=1/54

9. If all the letters of the word ‘TRIANGLE’ are taken and permuted in all possible ways, then the number of arrangements in which the relative positions of vowels and consonants remain unaltered is

(A) 180
(B) 720
(C) 360
(D) 540

10. The cities A, B, C, D of a state are located in such a way that they form the vertices of a regular polygon. Each city is connected to every other city by means of the shortest possible highway. The total number of highways connecting the cities is 28. What is the total number of ways in which a person can travel from the city A to city C, provided that during his travel he visits no city more than once.

(A) 1,296
(B) 127
(C) 1,957
(D) 762

2. Hello, sous!

8. A and B each throw a set of three dice.
If A throws a total of 16, B’s probability of getting a higher total is:

. . $\displaystyle (A)\;\frac{1}{54}\qquad (B)\;\frac{1}{36} \qquad (C)\;\frac{5}{108} \qquad (D)\;\frac{53}{54}$
There are: .$\displaystyle 6^3 = 216$ possible outcomes.

To get a higher total, $\displaystyle B$ must get a 17 or 18.

There are only 4 such outcomes: .$\displaystyle (5,6,6),\;(6,5,6),\;(6,6,5),\;(6,6,6)$

Therefore: .$\displaystyle P(B\text{ gets 17 or 18}) \;=\;\frac{4}{216} \;=\;\frac{1}{54}$

9. If all the letters of the word ‘TRIANGLE’ are taken and permuted in all ways,
then the number of arrangements in which the relative positions of vowels and consonants remain unaltered is:

. . $\displaystyle (A)\;180 \qquad (B)\;720 \qquad (C)\;360 \qquad (D)\;540$
Let: .$\displaystyle V$ - vowel, .$\displaystyle C$ = consonant.

The permutation has the form: .$\displaystyle CCVVCCCV$

The consonants can be placed in: .$\displaystyle 5!\,=\,120$ ways.

The vowels can be placed in: .$\displaystyle 3! \,=\,6$ ways.

Therefore, there are: .$\displaystyle 120\cdot6 \:=\:720$ permutations.

#10 has a typo.
I'll take a guess at what was meant.

10. The cities A, B, C, D, E, F, G, H are located so they form the vertices of a regular polygon.
Each city is connected to every other city by means of the shortest possible highway.
The total number of highways connecting the cities is 28.

What is the total number of ways in which a person can travel from the city A to city C,
provided that during his travel he visits no city more than once?

. . $\displaystyle (A)\; 1,\!296 \qquad (B)\;127 \qquad (C)\;1,\!957 \qquad (D)\; 762$
Since there are 28 connecting roads, there must be 8 cities.
I found no neat formula for this problem . . . I had to make a list.

One-step tour
He drives directly from $\displaystyle A$ to $\displaystyle C$: .1 way.

Two-step tour
He drives from $\displaystyle A$ to one of the other 6 cities,
. . then drives to $\displaystyle C$: .$\displaystyle _6P_1 \,=\,6$ ways.

Three-step tour
He drives from $\displaystyle A$ through two of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_2 \,=\,30$ ways.

Four-step tour
He drives from $\displaystyle A$ through three of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_3\,=\,120$ ways.

Five-step tour
He drives from $\displaystyle A$ through four of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_4 \,=\,360$ ways.

Six-step tour
He drives from $\displaystyle A$ through five of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_5 \,=\,720$ ways.

Seven-step tour
He drives from $\displaystyle A$ through all six of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_6 \,=\,720$ ways.

Therefore, there are: .$\displaystyle 1 + 6 + 30 + 120 + 360 + 720 + 720 \;=\;1957$ ways.

3. Originally Posted by Soroban
Hello, sous!

There are: .$\displaystyle 6^3 = 216$ possible outcomes.

To get a higher total, $\displaystyle B$ must get a 17 or 18.

There are only 4 such outcomes: .$\displaystyle (5,6,6),\;(6,5,6),\;(6,6,5),\;(6,6,6)$

Therefore: .$\displaystyle P(B\text{ gets 17 or 18}) \;=\;\frac{4}{216} \;=\;\frac{1}{54}$

Let: .$\displaystyle V$ - vowel, .$\displaystyle C$ = consonant.

The permutation has the form: .$\displaystyle CCVVCCCV$

The consonants can be placed in: .$\displaystyle 5!\,=\,120$ ways.

The vowels can be placed in: .$\displaystyle 3! \,=\,6$ ways.

Therefore, there are: .$\displaystyle 120\cdot6 \:=\:720$ permutations.

Since there are 28 connecting roads, there must be 8 cities.
I found no neat formula for this problem . . . I had to make a list.

One-step tour
He drives directly from $\displaystyle A$ to $\displaystyle C$: .1 way.

Two-step tour
He drives from $\displaystyle A$ to one of the other 6 cities,
. . then drives to $\displaystyle C$: .$\displaystyle _6P_1 \,=\,6$ ways.

Three-step tour
He drives from $\displaystyle A$ through two of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_2 \,=\,30$ ways.

Four-step tour
He drives from $\displaystyle A$ through three of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_3\,=\,120$ ways.

Five-step tour
He drives from $\displaystyle A$ through four of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_4 \,=\,360$ ways.

Six-step tour
He drives from $\displaystyle A$ through five of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_5 \,=\,720$ ways.

Seven-step tour
He drives from $\displaystyle A$ through all six of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_6 \,=\,720$ ways.

Therefore, there are: .$\displaystyle 1 + 6 + 30 + 120 + 360 + 720 + 720 \;=\;1957$ ways.
Thanks for your explaintion , I understood Q8 , Q9

for Q10 : I believe he said they are four cities only not 8 cities
plus if there will be total number of shortest high ways connecting all the cities = 28
that means 28 /4 = 7 ways ( for each city connecting other cities )
I guess there should be other ways than highways (normal ways to connect cities with each others)
Also , how can I make sure that "he visits no city more than once" in my calculations.
let me know what do you think

4. One-step tour
He drives directly from $\displaystyle A$ to $\displaystyle C$: .1 way.

Two-step tour
He drives from $\displaystyle A$ to one of the other 6 cities,
. . then drives to $\displaystyle C$: .$\displaystyle _6P_1 \,=\,6$ ways.

Three-step tour
He drives from $\displaystyle A$ through two of the other 6 cities (in some order)
. . then drives to $\displaystyle C$: .$\displaystyle _6P_2 \,=\,30$ ways.

[/quote]
Please confirm the rule : how you calculate [tex]_6P_2 \, ?
if n= 6 and K =2 then [tex]_6P_2 \, = n!/k!(n-k)!

5. Originally Posted by sous
Thanks for your explaintion , I understood Q8 , Q9

for Q10 : I believe he said they are four cities only not 8 cities
plus if there will be total number of shortest high ways connecting all the cities = 28
that means 28 /4 = 7 ways ( for each city connecting other cities )
I guess there should be other ways than highways (normal ways to connect cities with each others)
Also , how can I make sure that "he visits no city more than once" in my calculations.
let me know what do you think
If there are only four cities there can be only six shortest high ways connecting all the cities.