Probability Help PLEASE!! (2 Viewers)

[pLuvia]

Doesn't the identity factor play a role in circle permutations? Unlike arrangement in a line etc

 

[GaDaMIt]

Current Questions Requiring Help
Find how many arrangements of the letters of the word TRANSITION are possible if the Is, Ns and Ts are together?

I'm doing [5 x 6!/(2! x 2! x 2!)] x 4!.. and im getting 10800. Answer is 5400

Jesus. I don't remember how I did even that anymore? Someone care to explain this question to me. I'm revising the stuff atm =\

Like I don't get what the hell I did there, and no one really explained that to me seeing as I understood it at the time =\

 

[GaDaMIt]

Also

"By considering its prime factorisation, find the number of positive divisors of 315000."

I've got that down to 5^4 x 2^3 x 3^2 x 7.. dont get what to do from there?



AND

There are ten basketballers in a team. Find how many ways they can be split into two teams of five.

Wouldn't this be the same as how many ways they can make ONE team of five? As the rest would be left over and thus form the other team? Answer is given as half of that though (answer is 126) .. please explain?

 

[Riviet]

heres my 2 cents...
u do have the right expression, IE
[5! x 6!/(2! x 2! x 2!)], but you still have to halve it to get 5400 because of this: consider the letters in the 'block' I,I,N,N,T,T...we can have the permutation (INT)[INT], but that is identical to [INT](INT), where the 'INT' block switched places. Thus there is a duplicate of 'INT', not necessarily individual letters themselves.

Thus, halve your original answer to get the correct answer.

M.D has nicely explained it on the first page for you.

There are ten basketballers in a team. Find how many ways they can be split into two teams of five.

Wouldn't this be the same as how many ways they can make ONE team of five? As the rest would be left over and thus form the other team? Answer is given as half of that though (answer is 126) .. please explain?

Yes and yes. But you need to half your answer to get 126 because just like in the TRANSITION question, A1-A2-A3-A4-A5 and B1-B2-B3-B4-B5 is the same combination as choosing B1-B2-B3-B4-B5 and A1-A2-A3-A4-A5. So by taking 5C2, you have counted each combination twice, meaning you have to divide by two to get 126.

 

[Sober]

Also

"By considering its prime factorisation, find the number of positive divisors of 315000."

I've got that down to 5^4 x 2^3 x 3^2 x 7.. dont get what to do from there?

Factors:

5,5,5,5
2,2,2
3,3
7

Each divisor will be a multiple of several of these, that is we can shoose either 0,1,2,3, or 4 five's to include, etc.

Thus:

5x4x3x2 = 120 (this includes 1 and 315000)

 

[GaDaMIt]

A poker hand of five cards is dealt from a standard pack of 52. Find the probability of obtaining

a)one pair
b)two pairs
c)three of a kind
d)four of a kind
e)a full house (one pair and three of a kind)
f)a straight(five cards in sequence regardless of suit)
g)a flush(five cards of the same suit)
h)a royal flush(ten, jack, queen, king, and ace in a single suit)

I have no idea how to do any of them =\

 

[webby234]

Ok - I'll start with the last one (easier) and work backwards.

52C5 = 2 598 960 possible hands.

h) Four of these will be a royal flush (one for each suit) so 1/649 740

g) Ways to get five of a specific suit = 13C5 = 1287, four suits so multiply by four = 33/16 660

f) Get 2, 3, 4, 5, 6 in 4⁵ different ways (as there are four different cards of each kind). There are then 9 different starting places for the straight.

So 4⁵ * 9 = 192/54 145

 

[webby234]

e) Full house is 4C3 (ways of choosing three of the four twos for example) x 13 (possible three fo a kinds) x 4C2 (ways of choosing a pair of threes for exmaple) x 12 (number of possible two of a kinds remaining) = 3744

=6/4165

d) Four of a kind - 13 (different four of a kinds) * 48 (the other card) = 624

so 1/4165

 

[webby234]

c) Three of a kind - 13 * 4C3 * (48C2 - 4C2 * 12) (removing chance other one is pair) = 54912
= 88/4165

b) 4C2 * 4C2 * 44 * 13C2 = 123 552 (see (a) for my logic)

= 198/4165

a) Ways of getting a pair of twos with three other non twos - you can get the pair in 4C2 different ways (ie diamond + club etc.) you then get 3 of the other 48 cards. So

4C2 * (48C3 - 4C2 * 44 * 12) [removing chance of a second pair] * 13 (taking into account threes, fours etc.) Then remove the chance of getting two pairs and the chance of getting a full house, which include a pair.

= 1766/4165

There are probabbly simpler ways of doing these - such ways would be appreciated.

 

[GaDaMIt]

All correct except a) which is 352/833

 

[GaDaMIt]

There are two small rows of chairs, with 3 in the front row and 4 in the second row. How can 7 people be seated if:

i) 2 particular people must sit in the second row
ii) Jan will not sit on the same row as Norman and Stan must sit in the first row


explanation as well please?

Answers are 1440 and 1152

 

[Riviet]

i) 4P2 ways to arrange 2 people in the 4 seats of second row. There are 5 people left, so 5! ways to arrange them. So total no. of ways = 4P2.5! = 144.

ii) Let's say Jan sits at front and Norman sits at back. Also given Stan must sit at front, there are 3P2 ways to arrange Jan and Stan in the 3 possible seats of front row. 4P1 or 4! ways ot arrange Norman in the 4 seats of second row, and 4 people left to arrange, implying 4! ways for them to be arranged. We then multiply by 2 because instead, Jan could be sitting at back and Norman at front (in other words, they can swap places).
So total no. of way = 3P2.4P1.4!x2 = 1152.

 

[GaDaMIt]

A commuttee of 5 must be chosen from 8 boys and 7 girls. How many different committees can be formed if John will not go on the committee if Tracey is on the committe?

Answer = 2717

Explain please

 

[SoulSearcher]

Here, you have to calculate the number of committees able to be formed if Tracey is on the committee and John isnt, and the number of committees that John is on and Tracey isn't on.

Number of committees that Tracey is on and John isn't oM:
Since Tracey is on a committe, then John is not on the committee, therefore the number of people to choose from for boys and girls are reduced to 7 and 6 respectively. Therefore amount if committees:
7C4 + 7C3 * 6C1 + 7C2 * 6C2 + 7C1 * 6C3 + 6C4 = 715

Number of committees that John is on and Tracey isn't:
Since you can choose from any of the boys, but cannot choose Tracey, the amouunt of boys and girls you can choose from are 8 and 6 respectively. Therefore:
8C5 + 8C4 * 6C1 + 8C3 * 6C2 + 8C2 * 6C3 + 8C1 * 6C4 + 6C5 = 2002

Therefore total amount of committees = 715 + 2002 = 2717

 

[GaDaMIt]

I don't think all that was necessary. Don't think gender matters in this question ..

I've worked out an easy answer.. 15C5 - 13C3 = 2717

13C3 = amount of ways both are on table.. so take them off and yeah.. but i'm just looking for an alternative way to doing this question

 

[GaDaMIt]

Here, you have to calculate the number of committees able to be formed if Tracey is on the committee and John isnt, and the number of committees that John is on and Tracey isn't on.

Number of committees that Tracey is on and John isn't oM:
Since Tracey is on a committe, then John is not on the committee, therefore the number of people to choose from for boys and girls are reduced to 7 and 6 respectively. Therefore amount if committees:
7C4 + 7C3 * 6C1 + 7C2 * 6C2 + 7C1 * 6C3 + 6C4 = 715

Number of committees that John is on and Tracey isn't:
Since you can choose from any of the boys, but cannot choose Tracey, the amouunt of boys and girls you can choose from are 8 and 6 respectively. Therefore:
8C5 + 8C4 * 6C1 + 8C3 * 6C2 + 8C2 * 6C3 + 8C1 * 6C4 + 6C5 = 2002

Therefore total amount of committees = 715 + 2002 = 2717

Furthermore that doesn't include the possibility of them both not being on..

 

[Riviet]

Nice way of doing it, Gadamit, with the subtracting of undesired outcomes. Here's an alternate way of doing it:

Number of ways with Tracey on and John off = 13C4 = 715

Now if Tracey is off, John could be on OR off. So we have 14 people left to choose from, including John, since he can be on or off. So number of ways = 14C5 = 2002

Sum these two and you get your answer.