A Fistful of Brain Teasers

Don Atkinson posted:

Ok, a couple of straight forward probability tasters.

The first one inspired by those tubes of fruit pastilles...........

A box contains 2 red, 3 yellow and 5 green sweets. One is taken out at random and eaten. A second sweet is then taken out.

1. If the 1st sweet was green, what is the probability that the second sweet is also green ?

2. If the first sweet was not red, what is the probability that the 2nd sweet is red ?

Repeated here, otherwise it will probably get overlooked at the bottom of page 30..............

Don Atkinson posted:

The second one incorporates those proverbial fair dice..........(six sided cubes with sides numbered 1 to 6)

Two dice are thrown together.

1. What is the most likely score ?

2. what is the chance of getting this score three times in successive throws ?

1) 7

2) 1/216 (I think)

Don Atkinson posted:

Don Atkinson posted:

Ok, a couple of straight forward probability tasters.

The first one inspired by those tubes of fruit pastilles...........

A box contains 2 red, 3 yellow and 5 green sweets. One is taken out at random and eaten. A second sweet is then taken out.

1. If the 1st sweet was green, what is the probability that the second sweet is also green ?

2. If the first sweet was not red, what is the probability that the 2nd sweet is red ?

Repeated here, otherwise it will probably get overlooked at the bottom of page 30..............

I don’t think this is one where the first action influences the result so:

1st sweet green - , 2 red, 3 yellow, 4 green left, so chance of a green is 4/9.

1st sweet not red, 2 reds left in 9 sweets, so 2/9.

Eoink posted:

Don Atkinson posted:

The second one incorporates those proverbial fair dice..........(six sided cubes with sides numbered 1 to 6)

Two dice are thrown together.

1. What is the most likely score ?

2. what is the chance of getting this score three times in successive throws ?

1) 7

2) 1/216 (I think)

You should have more self-confidence Eoink !

Both are good answers !

Eoink posted:

Don Atkinson posted:

Don Atkinson posted:

Ok, a couple of straight forward probability tasters.

The first one inspired by those tubes of fruit pastilles...........

A box contains 2 red, 3 yellow and 5 green sweets. One is taken out at random and eaten. A second sweet is then taken out.

1. If the 1st sweet was green, what is the probability that the second sweet is also green ?

2. If the first sweet was not red, what is the probability that the 2nd sweet is red ?

Repeated here, otherwise it will probably get overlooked at the bottom of page 30..............

I don’t think this is one where the first action influences the result so:

1st sweet green - , 2 red, 3 yellow, 4 green left, so chance of a green is 4/9.

1st sweet not red, 2 reds left in 9 sweets, so 2/9.

 

You clearly know your traffic lights. Well Done !

This next one is probably (well, hopefully) a bit more challenging, but knowing you lot, you''l  all be thinking "really ?"....

Four letters are chosen at random from the word: DEALING.

What is the probability that :-

1. Exactly two vowels are chosen ?

2. At least two vowels are chosen ?

Don Atkinson posted:

This next one is probably (well, hopefully) a bit more challenging, but knowing you lot, you''l  all be thinking "really ?"....

Four letters are chosen at random from the word: DEALING.

What is the probability that :-

1. Exactly two vowels are chosen ?

2. At least two vowels are chosen ?

18/35, 22/35?

No need for the question mark Erich. Both are right !

Well done.

Looks like i’m Going to have to dig out some of my university maths book next !

Erich posted:

Don Atkinson posted:

This next one is probably (well, hopefully) a bit more challenging, but knowing you lot, you''l  all be thinking "really ?"....

Four letters are chosen at random from the word: DEALING.

What is the probability that :-

1. Exactly two vowels are chosen ?

2. At least two vowels are chosen ?

18/35, 22/35?

Hi Erich,

Could you provide the workings? I googled the theory, but failed to understand it all ...

rgds

My dad was a bookmaker. When throwing 2 dice, he would offer evens for under 7 or over 7, and 3 to 1 for the lucky seven, thereby ensuring that, on average, he should win.

ynwa250505 posted:

Erich posted:

Don Atkinson posted:

This next one is probably (well, hopefully) a bit more challenging, but knowing you lot, you''l  all be thinking "really ?"....

Four letters are chosen at random from the word: DEALING.

What is the probability that :-

1. Exactly two vowels are chosen ?

2. At least two vowels are chosen ?

18/35, 22/35?

Hi Erich,

Could you provide the workings?I googled the theory, but failed to understand it all ...

rgds

Googled........?????????

..... whatever happened to a "text book" ..... or am I out of touch in this modern (*) world...?

(*) Modern ? or perhaps that should be " modem"

Don Atkinson posted:

ynwa250505 posted:

Erich posted:

Don Atkinson posted:

This next one is probably (well, hopefully) a bit more challenging, but knowing you lot, you''l  all be thinking "really ?"....

Four letters are chosen at random from the word: DEALING.

What is the probability that :-

1. Exactly two vowels are chosen ?

2. At least two vowels are chosen ?

18/35, 22/35?

Hi Erich,

Could you provide the workings?I googled the theory, but failed to understand it all ...

rgds

Googled........?????????

..... whatever happened to a "text book" ..... or am I out of touch in this modern (*) world...?

(*) Modern ? or perhaps that should be " modem"

If I had the appropriate text book Don, I’d be all over it - much easier to flip pages than stare at screens!

What’s the weather in Vernon like atm?

Ah ! I kept virtually all of my school and university books. 

I’m back in the U.K., teaching.

Mrs D and the grandchildren tell me there’s plenty of snow up on Silver Star and the skiing is good. But school bekons ! And all will be back to normal this week.

I’ll give Erich or others a bit of time to explain the  18/35 and 22/35 before jumping in.

Total combination are C(7,4) = 7!/4!(7-4)! = 35

1. Combination for 3 vocals in groups of two * combinations for 4 consonants in groups of 2 / total combinations  = C(3,2) * C(4,2)/ 35 = 3!/2!(3-2)! * 4!/2!(4-2)!/35 = ( 6/2 * 24/4)/35 = 18/35

2. (Combinations for 3 vocals in groups of two  *  combinations for 4 consonants in groups of 2 + Combinations for 3 vocals in groups of 3 * combinations for 4 consonants in groups of 1) / total combinations = (18 (same result #1) + C(3,3)*C(4,1) )/35 = (18 + 1*4)/35 = 22/35

Sorry for the delay but I do have a different time and have to work a bit during weekdays.

Of course you can also do it by extension writing all the combinations. In this case only 35. And choose the good ones but it's easier (shorter)  with the mathematics.

Don Atkinson posted:

Ah ! I kept virtually all of my school and university books. 

I’m back in the U.K., teaching.

Mrs D and the grandchildren tell me there’s plenty of snow up on Silver Star and the skiing is good. But school bekons ! And all will be back to normal this week.hope the words above didn't imply

I’ll give Erich or othersa bit of time to explain the  18/35 and 22/35before jumping in.

Hi Erich,

Many thanks for taking the time to set out the detailed workings.

I hope the words above didn't imply "hurry up" or any similar "command" !! If so, I do apologise.

I am very conscious of time differences between the UK and the West Coast of Canada/USA etc and the fact that many (probably most) of us still have to work for a living ! I'm also conscious that it takes time to formulate text within the limitations of this Forum's host site and to find the time to set things out clearly. Many thanks for doing so.

Erich posted:

Total combination are C(7,4) = 7!/4!(7-4)! = 35

1. Combination for 3 vocals in groups of two * combinations for 4 consonants in groups of 2 / total combinations  = C(3,2) * C(4,2)/ 35 = 3!/2!(3-2)! * 4!/2!(4-2)!/35 = ( 6/2 * 24/4)/35 = 18/35

2. (Combinations for 3 vocals in groups of two  *  combinations for 4 consonants in groups of 2 + Combinations for 3 vocals in groups of 3 * combinations for 4 consonants in groups of 1) / total combinations = (18 (same result #1) + C(3,3)*C(4,1) )/35 = (18 + 1*4)/35 = 22/35

Sorry for the delay but I do have a different time and have to work a bit during weekdays.

Many thanks Erich - much appreciated! I'll be looking for an introductory probability text on my visit to a bookstore ...

ynwa250505 posted:

Erich posted:

Total combination are C(7,4) = 7!/4!(7-4)! = 35

1. Combination for 3 vocals in groups of two * combinations for 4 consonants in groups of 2 / total combinations  = C(3,2) * C(4,2)/ 35 = 3!/2!(3-2)! * 4!/2!(4-2)!/35 = ( 6/2 * 24/4)/35 = 18/35

2. (Combinations for 3 vocals in groups of two  *  combinations for 4 consonants in groups of 2 + Combinations for 3 vocals in groups of 3 * combinations for 4 consonants in groups of 1) / total combinations = (18 (same result #1) + C(3,3)*C(4,1) )/35 = (18 + 1*4)/35 = 22/35

Sorry for the delay but I do have a different time and have to work a bit during weekdays.

Many thanks Erich - much appreciated! I'll be looking for an introductory probability text on my visit to a bookstore ...

Hi YNWA,

Make sure that the book also covers Permutations and Combinations.

They aren’t absolutely necessary, but they are very helpful when dealing with some probability issues.

The books that I current use were written by Bostock and Chandler for A Levels.

Don Atkinson posted:

ynwa250505 posted:

Erich posted:

Total combination are C(7,4) = 7!/4!(7-4)! = 35

1. Combination for 3 vocals in groups of two * combinations for 4 consonants in groups of 2 / total combinations  = C(3,2) * C(4,2)/ 35 = 3!/2!(3-2)! * 4!/2!(4-2)!/35 = ( 6/2 * 24/4)/35 = 18/35

2. (Combinations for 3 vocals in groups of two  *  combinations for 4 consonants in groups of 2 + Combinations for 3 vocals in groups of 3 * combinations for 4 consonants in groups of 1) / total combinations = (18 (same result #1) + C(3,3)*C(4,1) )/35 = (18 + 1*4)/35 = 22/35

Sorry for the delay but I do have a different time and have to work a bit during weekdays.

Many thanks Erich - much appreciated! I'll be looking for an introductory probability text on my visit to a bookstore ...

Hi YNWA,

Make sure that the book also covers Permutations and Combinations.

They aren’t absolutely necessary, but they are very helpful when dealing with some probability issues.

The books that I current use were written by Bostock and Chandler for A Levels.

 

Don Atkinson posted:

ynwa250505 posted:

Erich posted:

Total combination are C(7,4) = 7!/4!(7-4)! = 35

1. Combination for 3 vocals in groups of two * combinations for 4 consonants in groups of 2 / total combinations  = C(3,2) * C(4,2)/ 35 = 3!/2!(3-2)! * 4!/2!(4-2)!/35 = ( 6/2 * 24/4)/35 = 18/35

2. (Combinations for 3 vocals in groups of two  *  combinations for 4 consonants in groups of 2 + Combinations for 3 vocals in groups of 3 * combinations for 4 consonants in groups of 1) / total combinations = (18 (same result #1) + C(3,3)*C(4,1) )/35 = (18 + 1*4)/35 = 22/35

Sorry for the delay but I do have a different time and have to work a bit during weekdays.

Many thanks Erich - much appreciated! I'll be looking for an introductory probability text on my visit to a bookstore ...

Hi YNWA,

Make sure that the book also covers Permutations and Combinations.

They aren’t absolutely necessary, but they are very helpful when dealing with some probability issues.

The books that I current use were written by Bostock and Chandler for A Levels.

Noted and thanks

A permutation is an ordered arrangement of a number of items.

A combination is an unordered selection of a number of items from a given set.

In each of the following problems determine, without  working out the answer, whether you are asked to find a number of permutations, or a number of combinations.

1. How many arrangements of the letters A, B, C are there ?

2. A Liverpool FC team of eleven members can be selected from a squad of eighteen players. How many different teams can be selected ?

3. A Naim Forumite can take eight records to a desert island, chosen from his own collection of one hundred records. How many different sets of records could he choose ?

4. The first, second and third prizes for the Naim raffle are awarded by drawing tickets from a box of five hundred. In how many ways can the prizes be won ?

5. A Salisbury telephone number is a seven digit number. How many Salisbury telephone lines are available ?

6. One red dice and one green dice are rolled (each numbered one to six). In how many ways can a total score of six be obtained ?

Don Atkinson posted:

Don Atkinson posted:

Ah ! I kept virtually all of my school and university books. 

I’m back in the U.K., teaching.

Mrs D and the grandchildren tell me there’s plenty of snow up on Silver Star and the skiing is good. But school bekons ! And all will be back to normal this week.hope the words above didn't imply

I’ll give Erich or othersa bit of time to explain the  18/35 and 22/35before jumping in.

Hi Erich,

Many thanks for taking the time to set out the detailed workings.

I hope the words above didn't imply "hurry up" or any similar "command" !! If so, I do apologise.

I am very conscious of time differences between the UK and the West Coast of Canada/USA etc and the fact that many (probably most) of us still have to work for a living ! I'm also conscious that it takes time to formulate text within the limitations of this Forum's host site and to find the time to set things out clearly. Many thanks for doing so.

No need to apologise in any way. I should have published the math in the same post. Best regards.

Permutations or Combinations ?

C,C,C,P,P,C  

Hi YNWA,

I think a visit to Waterstones is in order ............. (half of them were correct)

Consider the following situation:

A street vendor stocks ten weekly periodicals and has a display stand with five racks in a vertical column. He clearly cannot display all ten of his magazines so first he must choose a group of five. The order in which he picks up his chosen periodicals is irrelevant: the set of five is only one combination. But there are quite a few possible sets ie combinations.

Once he has made his choice (of combination) he is then able to place the five periodicals in various different orders on the display stand. He is now arranging them and each possible arrangement is a permutation.

ie a particular set of five publications is one combination, but can be arranged to give several different permutations.

PCCCPP?

Dozey posted:

PCCCPP?

5/6.

Cheers, Don