Problems on Probability

Prakash Panneer · 2006-12-20 11:06:06

Hi all,

Can anyone help me with these problems?

Please!!!!!!!!!!!!!!!!!!

1. In a single throw of two dice, what is the probability that neither a doublet nor a 7 will appear?

2.Seven dice are rolled. Calling a 5 or 6 a success, find the probability of getting
(a) exactly 4 successes;
(b) at most 4 successes.

3. If a die is thrown repeatedly, what is the probability that a 6 is obtained before a 1 turns up?

4.A coin is tossed repeatedly. Find the probability that the fourth head appears on the eleventh toss.

5.Four statisticians arrange to meet at the Grand Hotel in Paris. It happens that there are 4 hotels with that name in the city. What is the probability that all the statisticians will choose different hotels?

6.A family has 6 children. Find the probability that there are fewer boys than girls. Assume that the probability of the birth of either gender is ½.

Thanks in advance

John E. Franklin · 2006-12-20 15:52:33

Can you keep these questions separate, it just gets all messy when lots of people try to go off on tangents about each individual question. Sorry to be so picky (smile)

pi man · 2006-12-20 16:56:19

1. No doublet or 7's are thrown. There are 36 (6*6) different possible throws. There are 6 doubles (11, 22, 33, 44, 55, 66) and 6 ways to roll a 7 (16, 25, 34, 43, 52, 61). None of them are duplicated, so there are 24 ways (36-6-6) out of the 36 ways to not throw a doublet or 7. Thats .666 or 66.6%

John E. Franklin · 2006-12-20 18:06:48

2.) one-third chance of rolling a 5 or a 6 on any given die and 2/3's chance of rolling a 1,2,3,4 on any given die.
So hence, for seven dice, and four at 1/3 chance, you get:
(1/3) x (1/3) x (1/3) x (1/3) x (2/3) x (2/3) x (2/3), and this occurs in other orders too.l
Such as
(and call 1/3, just 3 and call 2/3's just 6 for short)
So now, we get:
3333666
3336366
3336636
3336663
3363366
3363636
3363663
3633366
3633636
3633663
3636336
3636363
oh what's next, hmmm...
3636633
3663336
3663363
3663633
3666333
6333366
6333636
6333 I must be ill doing this!! Right???
6333663 (all of them have 4 threes in case you didn't notice)
(We need to know how many there are with 4 threes since I don't know the formula for that.)
6336336
6336363
6336633
6363336
6363363
6363633
6366333 Gee this just gets easier and easier the more you do it!!!
6633336
6633363
6633633
6636333
6663333 YeeeeHaaaa!!!!! Time to count them up.

John E. Franklin · 2006-12-20 18:14:57

0.117055327~ is approximately answer for 2a), from 32 times (1/3) up 4 times (2/3) up 3.
Hope it's right!!!! Can't be sure.

John E. Franklin · 2006-12-20 18:29:53

For 2b) I'm gonna take a wild stab at it.
We want probability for 0,1,2,3,4 all combined.
Hence some examples are 6666666, no successes.
6636666 is an example of one success, for which there are seven of these.
6636366 is an example of two successes, for which there are maybe 24??? of these, I have no idea without
going thru them. Anyone know the formula?? And what is the formula called?
I think it is 6 + 5 + 4 + 3 + 2 + 1 ?, which is 21, hmm, not sure, getting to late to work on this.
Bye. Good Luck.

mathsyperson · 2006-12-21 02:24:14

I hate to break it to you John, but there you missed out 3 combinations in your big list.
3366336
3366363
3366633

So there are 35 combinations in all, which makes the answer to 2a) become 35*(1/3)^4*(2/3)^3 = 280/2187 ≈ 0.128

You've made a good start on the second question, considering you were working without the very useful formula that you wanted.

As you said, there is 1 way for there to be no successes, there are 7 ways for there to be 1 success and you guessed correctly that there are 21 ways for there to be 2 successes.
There are 35 ways for there to be 3, because it is identical to the situation of there being 4, just with the 3's and 6's switched.

And then to work out the total probability you'd just need to add together each of the individual ones.

---

3) This question is trivial. If you keep throwing the die until you get a 1 or 6, then none of the other sides of the die are important, because they all just mean that you throw it again. Therefore, you might as well just toss a coin labelled 1 and 6 rather than heads and tails. And I'm sure you know the probabilities of tossing a coin.

John E. Franklin · 2006-12-21 03:52:56

Yeah mathsy, that's why I knew I was off by 3 somewhere as I said 21 or 24, somethings wrong because I knew they all added up to 2^7 or 128, and something was missing. Thanks for finding them, I just found them too redoing the list in a different order.
By the way, what is this formula for 1,7,21,35?? Is it 7x0, 7x1, 7x3, 7x5 ?? Or is that a coincidence??
Also, if anyone has a neat order you can put all 35 combinations in besides ascending order, what would be another order that
brings inspiration into permutation ideas??

Stanley_Marsh · 2006-12-21 04:18:44

4.everytime the coin is tossed the probability is 1/2 , the fourth head is fixed , So there are 3 heads in the early toss,that's like 10 place for 3 heads ,there are C10(3) kinds of arragnement , then 4 heads , 7 tails ,we get (1/2)^4*(1/2)^7 . I think the probability is C10(3) *(1/2)^4*(1/2)^7 = 0.05859375 , Dont know if I am correct

Stanley_Marsh · 2006-12-21 04:27:09

5. statisticians A,B,C,D , Hotel W,X, Y ,Z

Assume A picks W , we have [BX,CY,DZ] [BY,CX,DZ] ......... 6 combination
Since A can choose between W,X,Y,Z , ,so there will be 6x4 =24 combinations
That's the situation when everyone chooses different hotel.

Consider all the situations , 4 can choose one hotel , maybe 3 , 2 ,1 .. so A may choose W,X ,Y,Z, since each of them has 4 choices , the sum of the situation is 4x4x4x4 =256

the probability p=24/256 =0.09375

Stanley_Marsh · 2006-12-21 04:33:27

6. Denote boy as B , girl as G, when there are fewer boys than girls , the satisfying condition is 2B4G, 1B5G, 0B6G
Every birth can be boy or girl , so there are 2^6 combinations
Consider the 2B4G condition , due to the differents order of the birth , there are C6(2) combination
Consider the 1B5G condition , there are 6 combination
Consider the 0B6G condition , there is only one
the probabilty p= [C6(2)+6+1] /[2^6]

mathsyperson · 2006-12-21 06:21:56

John E. Franklin wrote:

Yeah mathsy, that's why I knew I was off by 3 somewhere as I said 21 or 24, somethings wrong because I knew they all added up to 2^7 or 128, and something was missing. Thanks for finding them, I just found them too redoing the list in a different order.
By the way, what is this formula for 1,7,21,35?? Is it 7x0, 7x1, 7x3, 7x5 ?? Or is that a coincidence??
Also, if anyone has a neat order you can put all 35 combinations in besides ascending order, what would be another order that
brings inspiration into permutation ideas??

Oops, I meant to give you the formula in my previous post, but I forgot. Anyway, if there are "n" trials and "r" successes, then the total number of ways that that can happen is:

n!
------------
r! * (n-r)!

I think the pattern you noticed is just a coincidence. For example, if there were 4 trials then there would be 1, 4, 6, 4 and 1 combinations for 0, 1, 2, 3 and 4 successes respectively, and the 6 is not a multiple of 4, so the pattern doesn't always hold. Plus it never works for the 1's.

Prakash Panneer · 2006-12-21 06:50:51

Thanks a lot

John E. Franklin · 2006-12-21 11:08:31

Your welcome!
Thanks a zillion, mathsy!!!
n!
------------
r! * (n-r)!
now means a lot more to me since I've done some examples today.
Today I did (1 + 9 + 36 + 84 + 126 )2 = 2^9 = 512, for n=9.
I didn't write down all 512, but I wrote down them by certain rotating groups and mirrors.
For example, the rol and ror assembly language routines on 8 bits in a byte rotated.
Except I was working with 9bits on paper.

What is this formula called?? Binomial probability ?? Binary Choices ?

Last edited by John E. Franklin (2006-12-21 11:19:48)

George,Y · 2006-12-21 15:34:06

Stanely_Marsh uses the standard way.

4) First 6[sup]6[/sup] possible basic outcomes(ways), and each outcome share the same probability of 1/6[sup]6[/sup]. Agree?

Now we need only to count how many basic outcomes coincide with a 6 before a 1.

Excluding the ways without 6 or 1, we can substract this number from the total number of ways:
4[sup]6[/sup]

Because of the symmetry of 6 vs 1, we can predict the ways of 6-1 and 1-6 are the same. Strictly, whenever there is a way of 6-1, there is definately a way of 1-6 created by swapping the two, vise-versa. So they share the same amount of ways.

Thus for 1-6 or 6-1
there are
(6[sup]6[/sup]-4[sup]6[/sup])/2 ways.

And each way has the probability of 1/6[sup]6[/sup]

So the possiblity of all the ways satisfying 1-6 or 6-1 together should be
(6[sup]6[/sup]-4[sup]6[/sup])/2 (1/6[sup]6[/sup])

Binary choices. It can never be binomial probability because it is acctually larger than 1.

Math Is Fun Forum

#1 2006-12-20 11:06:06

Problems on Probability

#2 2006-12-20 15:52:33

Re: Problems on Probability

#3 2006-12-20 16:56:19

Re: Problems on Probability

#4 2006-12-20 18:06:48

Re: Problems on Probability

#5 2006-12-20 18:14:57

Re: Problems on Probability

#6 2006-12-20 18:29:53

Re: Problems on Probability

#7 2006-12-21 02:24:14

Re: Problems on Probability

#8 2006-12-21 03:52:56

Re: Problems on Probability

#9 2006-12-21 04:18:44

Re: Problems on Probability

#10 2006-12-21 04:27:09

Re: Problems on Probability

#11 2006-12-21 04:33:27

Re: Problems on Probability

#12 2006-12-21 06:21:56

Re: Problems on Probability

#13 2006-12-21 06:50:51

Re: Problems on Probability

#14 2006-12-21 11:08:31

Re: Problems on Probability

#15 2006-12-21 15:34:06

Re: Problems on Probability

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