Permutations/Combinations Question (1 Viewer)

[needpapers]

From exercise 10H of Cambridge Q 22.(a)

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

Thanks.

 

[-tal-]

I got for
a) 26
b) 26C2

 

[u-borat]

i'd appreciate it if someone cuold answer this as well.

 

[Enteebee]

http://www.cs.uvm.edu/~cpratt/solutions/Homework3.html

 

[u-borat]

thanks a lot ntb

 

[needpapers]

Thanks a lot Enteebee. I have another question from Exercise 10I - Q24
(a) Five diners in a restaurant choose randomly from a menu featuring five main courses. Find the probability that exactly one of the main courses is not chosen by any of the diners
(b) Repeat the question if there are n diners and a choice of n main courses.

Thanks.

 

[bored of sc]

Thanks a lot Enteebee. I have another question from Exercise 10I - Q24
(a) Five diners in a restaurant choose randomly from a menu featuring five main courses. Find the probability that exactly one of the main courses is not chosen by any of the diners.
(b) Repeat the question if there are n diners and a choice of n main courses.

Thanks.

(a) 3125 ways in which the courses can be chosen i.e. (5C1)⁵
4C1*(5C1)⁴*5 ways in which one of the main courses is not chosen by any person = 100
100/3125 = 4/125 --- I hope this is right.

(b) [nC1*(nC1)^n-1*n]/(nC1)ⁿ

*Crosses fingers.*

Maths is so hard.

 

[u-borat]

answer is 48/125 for part i) according to the answers.

 

[lolokay]

a) there is 5⁵ = 3125 total ways

say the first person chooses a meal, the second a different meal, the 3rd a different meal again, the 4th a different meal, and the 5th chooses a meal already chosen

this is done in 5*4*3*2*4 ways = 480
now looking at the last person (the only one who needs to really be considered with ordering): he can be put any where amongst the other 4, so multiply the answer by 5. Then, divide by 2 since 2 people have the same meal

this gives 480/3125 *5/2 = 48/125

b) the general case:
n!*(n-1)n/(2*nⁿ) based on my reasoning for a)



the question at the start of the thread is just..

there are 52!/47!5! total ways of dealing 5 cards

for a pair, there are 13 values to take a pair from. There are 6 ways of choosing the suits from this value. The remaining 3 cards must be of different values, ie. 12*4*11*4*10*4 for the ways of choosing them, 3! to unorder the 3 non-pairs
this gives P(pair) = 352/833

for choosing 2 pairs, it's 13*12 (to choose the pair values) *6*6 (to choose the suits of the values) /2 to unorder the 2 pairs from each other * 44 to choose the remaining card
gives a probability of 198/4165

 

[needpapers]

a) there is 5⁵ = 3125 total ways

say the first person chooses a meal, the second a different meal, the 3rd a different meal again, the 4th a different meal, and the 5th chooses a meal already chosen

this is done in 5*4*3*2*4 ways = 480
now looking at the last person (the only one who needs to really be considered with ordering): he can be put any where amongst the other 4, so multiply the answer by 5. Then, divide by 2 since 2 people have the same meal

this gives 480/3125 *5/2 = 48/125

Good work, but I don't understand why they have to choose different meals. Why can't they all choose the same meal or multiple people choose a particular meal also?

 

[lolokay]

then you would have more than one meal that isn't chosen, so there has to be 4 that are different from each other, and one that is the same as one of the other 4

 

[bored of sc]

a) there is 5⁵ = 3125 total ways

say the first person chooses a meal, the second a different meal, the 3rd a different meal again, the 4th a different meal, and the 5th chooses a meal already chosen

this is done in 5*4*3*2*4 ways = 480
now looking at the last person (the only one who needs to really be considered with ordering): he can be put any where amongst the other 4, so multiply the answer by 5. Then, divide by 2 since 2 people have the same meal

this gives 480/3125 *5/2 = 48/125

b) the general case:
n!*(n-1)n/(2*nⁿ) based on my reasoning for a)



the question at the start of the thread is just..

there are 52!/47!5! total ways of dealing 5 cards

for a pair, there are 13 values to take a pair from. There are 6 ways of choosing the suits from this value. The remaining 3 cards must be of different values, ie. 12*4*11*4*10*4 for the ways of choosing them, 3! to unorder the 3 non-pairs
this gives P(pair) = 352/833

for choosing 2 pairs, it's 13*12 (to choose the pair values) *6*6 (to choose the suits of the values) /2 to unorder the 2 pairs from each other * 44 to choose the remaining card
gives a probability of 198/4165

Give me your brain - NOW!

 

[sam168]

my solution is different but not mach the answer

assume one course in nobody choose so the first dinner have 4 choise, 2nd have 3, 3rd have 2, 4th have 1 and 5th have 4, total will be 4!.
since there are 5 courses so the numbers will be 4!x5/2! why mine is wrong?

 

[sam168]

answer is 48/125 for part i) according to the answers.

my solution is different but not mach the answer

assume one course in nobody choose so the first dinner have 4 choise, 2nd have 3, 3rd have 2, 4th have 1 and 5th have 4, total will be 4!.
since there are 5 courses so the numbers will be 4!x5/2! why mine is wrong?

 

[braintic]

my solution is different but not mach the answer

assume one course in nobody choose so the first dinner have 4 choise, 2nd have 3, 3rd have 2, 4th have 1 and 5th have 4, total will be 4!.
since there are 5 courses so the numbers will be 4!x5/2! why mine is wrong?

You are making the 5th person different to the others.
In your solution, the 5th person can have the same course as ANY of the other 4 people, yet the 4th person cannot have the same course as the 1st, 2nd or 3rd person.