Probability Help PLEASE!! (1 Viewer)

[GaDaMIt]

I do not follow your reasoning, here is how I figured it, we look at all the ways that violate the conditions, those of the form of how many pairs are together:

AABBCCDD (4 pairs) = 4! = 24
AA?BB?CC (3 pairs) = 5!/2 - 24 = 36
?AA??BB? (2 pairs) = 6!/4 - 24 - 36x2 = 84
???AA??? (1 pair) = 6!/8 - 24 - 36x3 - 84x3 = 984

Then we sum each of these results by the ways they can happen using different orders of pairs = 24 + 4x36 + 6x84 + 984 = 1656, this subtracted by all the possible arrangements (8!/16 = 2520) is 864.

This probably makes absolutely no sense to anybody and is probably far more convoluted than the previous solution but I am pretty sure the question is beyond the scope of a highschool probability question.

Well all my questions are from the Cambridge year 12 3u book .. and yeh.. that Bob question was an extension question heh

 

[Riviet]

How many ways can the letters of the word CONNECTION be arranged if all the N's are together? Answer = 10 080 Working is 8! / (2! x 2!) .. i dont get why these numbers though.. anyone care to enlighten me?

Edit: Reason i dont get this is that the 3 Ns can be together so I'm thinking 3Ns fitting into a 10 letter slot thing being together can happen 8 ways..

So im thinking like 7! (2! x 2!) x 8

EDIT: LOL that gives the same answer.. ok so im getting it right -.- anyway.. can someone tell me why they have that working instead of my working?

Given that the N's are together, consider the 3 N's as ONE unit. Therefore, we have 8 different units to arrange, ie the N's as 1 unit PLUS the other 7. Therefore total number of arrangemnts = 8!. But we two of the other 7 letters are doubles, ie we have 2 O's and 2 C's so we have to divide by 2! twice, because OO is the same thing as OO and CC is the same as CC.

 

[GaDaMIt]

yeah i get it now.. they just joined the 7! x 8 to make 8!..

another question

from a standard pack of 52 cards, three are selected at random. find the probability that:

a) they are the jack of spades, two of clubs and the seven of diamonds ANSWERED

I'm thinking about this logically, so why isn't 1 / 52 x 1 / 51 x 1 / 50 correct? That gives 1/132600. Answer is 1/22100..

Edit: Further questions

i) exactly one is a diamond (no idea how to do..)

j) at least two of them are diamonds

7) Three boys and three girls are to sit in a row. Find the probability that:
c) two specific girls sit next to one another

 

[Riviet]

a) Remember that "drawing the jack of spades, then the two of clubs and then the seven of diamonds" is just 1 of the 3! ways of arranging. So there is a 6 x 1/132600 chance of drawing these three cards in any order.

Stay tuned for more. XD

 

[Riviet]

(i) Split the deck into 2: all the diamands (13 cards) in a pile and all the non diamands (39 cards) in another pile. Let's say we want the first two to be non diamand and the third to be the diamand card. For the first, there would be a 39/52 chance of choosing a non-diamand. For the second card, there would be a 38/51 chance. For the third, it must be a diamand to match the question's condition of having exactly 1 diamand. So there's a 13/50 chance of choosing one. However, there are 3! ways that we can choose these three cards, so we must multiply the three fractions by 3!:
39/52 x 38/51 x 13/50 x 3! = 741/850.

(j) We consider 2 cases:
2 diamands and 1 non-diamand: 13/52 x 12/51 x 39/50 x 3! = 117/425

3 diamands and 0 non-diamands: 13/52 x 12/51 x 11/51 x 3! = 22/289

Add these two probabilities since they are separate cases = 2539/7225.

 

[Riviet]

7) Three boys and three girls are to sit in a row. Find the probability that:
c) two specific girls sit next to one another

Total number of unrestricted ways = 6! = 720
We want to find the number of ways that 2 specific girls sit adjacent each other. Consider them as 1 unit. There are 3 other guys and 1 other girl left, which makes a total of 5 units. There are 5! ways to arrange these as well as 2! ways to arrange the two girls that are [stuck] together.
So total number of ways = 5! x 2! = 240
.'. probability = 240/720 = 1/3.

 

[GaDaMIt]

(i) Split the deck into 2: all the diamands (13 cards) in a pile and all the non diamands (39 cards) in another pile. Let's say we want the first two to be non diamand and the third to be the diamand card. For the first, there would be a 39/52 chance of choosing a non-diamand. For the second card, there would be a 38/51 chance. For the third, it must be a diamand to match the question's condition of having exactly 1 diamand. So there's a 13/50 chance of choosing one. However, there are 3! ways that we can choose these three cards, so we must multiply the three fractions by 3!:
39/52 x 38/51 x 13/50 x 3! = 741/850.

(j) We consider 2 cases:
2 diamands and 1 non-diamand: 13/52 x 12/51 x 39/50 x 3! = 117/425

3 diamands and 0 non-diamands: 13/52 x 12/51 x 11/51 x 3! = 22/289

Add these two probabilities since they are separate cases = 2539/7225.

ahh you got those wrong.. i) 741/1700 and j) 64/425

 

[Riviet]

Damn, already rusty after a week of holidays :\

I'll have another go:

i) Let's just say we want the 2 non-diamand cards first with the diamand card last:
First two cards: ³⁹C₂ ways of drawing
Third card: ¹³C₁ ways of drawing
Since, this is happening simultaneously, multiply them together. However, this is only 1 way of drawing the three cards so we multiply by 3! since there are 3! ways of drawing these 3 cards.
So number of ways = ³⁹C₂.¹³C₁.3!=57798.
Total number of ways = 52x51x50 = 132600.
.'. probability = 57798/132600 = 741/1700

EDIT: revised answer.

 

[webby234]

3 diamonds - 13C3 (ways of choosing three of the thirteen diamonds)
2 diamonds - 13C2 * 39 (ways of choosing two diamonds and one other card)

Total ways - 52C3

(13C2 * 39 + 13C3)/52C3

= 64/425

 

[pLuvia]

(j)
At least two diamonds
2 diamonds and 1 non-diamond - ¹³C₂*³⁹C₁=3024
3 diamonds and 0 non-diamond - ¹³C₃*³⁹C₀=286

Since both are ways that it can happen, add both together
Total number of ways = ⁵²C₃=22100
Probability = 3328/22100=64/425

Edit: Webby beat me to it

 

[GaDaMIt]

Damn, already rusty after a week of holidays :\

I'll have another go:

i) Let's just say we want the 2 non-diamand cards first with the diamand card last:
First card: ³⁹C₁ ways of drawing
Second card: ³⁸C₁ ways of drawing
Third card: ¹³C₁ ways of drawing
Since, this is happening simultaneously, multiply together. This is only 1 way of drawing the three cards so we multiply by 3 since the diamand card can be drawn first, second or third (3 ways).
So number of ways = ³⁹C₁.³⁸C₁.¹³C₁.3=57798.
Total number of ways = 52x51x50 = 132600.
.'. probability = 57798/13260 = 741/1700

why is denominator here 52P3 instead of 52C3. For all the other questions in this set of questions the denominator is 52C3...

 

[webby234]

He's just done it a different way. He's considered what order you draw them in, then subtracted the combinations that are the same but in a different order.

I like my way better

 

[GaDaMIt]

He's just done it a different way. He's considered what order you draw them in, then subtracted the combinations that are the same but in a different order.

I like my way better

could you do i) in your way then?

 

[pLuvia]

from a standard pack of 52 cards, three are selected at random. find the probability that:

a) they are the jack of spades, two of clubs and the seven of diamonds ANSWERED

I'm thinking about this logically, so why isn't 1 / 52 x 1 / 51 x 1 / 50 correct? That gives 1/132600. Answer is 1/22100..

From observation that selection can only happen in one way, and the total number of ways three cards could be drawn is ⁵²C₃. Hence the answer is 1/22100

 

[Riviet]

Revised my answer for (i) in my last post.

 

[GaDaMIt]

Arhh another question.. im doing something wrong and i see what it is but i dont know why its wrong.

The digits 3,3,4,4,4 and 5 are placed in a row to form a six-digit number. If one of these numbers is selected at random, find the probability that:
d) the number starts with 5 and then the 4s and 3s alternate.

Im doing 1/6 (coz theres a 1/6 chance of being a 5 for first number) x 3C1 x 2C1 x 2C1 x 1C1 x 1C1 divided by 6P6

So to make it clearer

1/6 x 3! x 2!
---------------
6P6

= 2 / 720 = 1/360

Answer is just simply 1/60 so I see that im not supposed to multiply with the 1/6.. can someone explain why to me please?

 

[Riviet]

Your method is perfectly correct, just your last line where you have the 2!, 1C1.1C1 is not equal to 2!. Just a careless mistake. You should get the right answer now.

 

[Riviet]

Ah sorry, my bad. Didn't see the two 2C1's. Anway, it's probably best to just calculate the number of ways first before worrying about the probability. You should follow the two step method that webby has outlined above.

 

[GaDaMIt]

13) The letters of KETTLE are arranged randomly in a row. Find the probability that:

c) the two letters E are together and the two letters T are together

no idea =\ lol.. asking so many probability questions .. argh .. but i feel so much better i can answer every question in the past 3u maths tests aside from the last one on both which is about people sitting a circle thats my next exercise.. not up to it yet ..

 

[Riviet]

Number of ways with the T's together and E's together = 4! = 24
Total number of unrestricted ways = 6!/(2!x2!) = 180
.'. probability = 24/180 = 2/15.

Note: I am assuming that the pairs of letters are counted as the same when swapped around.