Probability help (1 Viewer)

[carolinevivi]

Please help me with these questions:

1) number more than 40000 are formed from a digits 1,3,4,5,7 and 9 without repetition. How many of them are odd?

2) 3 girls and 3 boys are to be seated at a round table. How many seating arrangements are possible if each of the boy is to have a girl seated either side of them?

3) Couple and their 5 children go to the movie. what are the probability that the five children are located seats next to each other ?.

4) In the word DEMAND how many of the words will the D's be separated ?

Many thanks,

 

[obliviousninja]

I'll have a look after my gaming sesh. 30min.

 

[obliviousninja]

1. If larger than 400000, first digit must be 4,5,7,9

Taking the case where 4 is the first digit. 1 _ _ _ _ 5 (5 possibilities for last digit to be off) So. 4P4 x 5

Taking the case where 5,7,9 is the first digit 3 _ _ _ _ 4 So. 3 x 4P4 x 4

2. Seat the first boy. Then if there has to be girls either side, seat one of the 3 girls and one of the 2 remaining. There is one remaining position next to these 2 girls respectively. And the remaining seats are filled. So 3 x 2 x 2

3. Two ways to arrange the groups, the couple and the five. Within the couple, this can be arranged in 2! ways. Within the group, 5! ways. So 2 x 2! x 5!

4. Just consider each case

 

[RealiseNothing]

1. If larger than 400000, first digit must be 4,5,7,9

Taking the case where 4 is the first digit. 1 _ _ _ _ 5 (5 possibilities for last digit to be off) So. 4P4 x 5

Taking the case where 5,7,9 is the first digit 3 _ _ _ _ 4 So. 3 x 4P4 x 4

2. Seat the first boy. Then if there has to be girls either side, seat one of the 3 girls and one of the 2 remaining. There is one remaining position next to these 2 girls respectively. And the remaining seats are filled. So 3 x 2 x 2

3. Two ways to arrange the groups, the couple and the five. Within the couple, this can be arranged in 2! ways. Within the group, 5! ways. So 2 x 2! x 5!

Sick permutations in 2U.

 

[anomalousdecay]

Sick permutations in 2U.

IKR

Please help me with these questions:

1) number more than 40000 are formed from a digits 1,3,4,5,7 and 9 without repetition. How many of them are odd?

2) 3 girls and 3 boys are to be seated at a round table. How many seating arrangements are possible if each of the boy is to have a girl seated either side of them?

3) Couple and their 5 children go to the movie. what are the probability that the five children are located seats next to each other ?.

4) In the word DEMAND how many of the words will the D's be separated ?

Many thanks,

1) the first digit must be either 4, 5, 7 or 9.

So the probability. So if it is four, then the possibility of odd numbers is:


This takes into account that all the remaining digits will produce an odd number.

If the number starts with the digit 5, then the possibility of odd numbers must account for the fact that 4 will produce an even number:




To get the total odd possibilities, the case above will occur if the number starts with either digits 7 or 9 as well:



2)

Seat the first boy. Then if there has to be girls either side, seat one of the 3 girls and one of the 2 remaining. There is one remaining position next to these 2 girls respectively. And the remaining seats are filled. So 3 x 2 x 2

3) Using 3-unit method I got:

I'm not sure how you do it by 2-unit method.

4) Using 3-unit method, we got the total outcomes of the word:



With the words together, we get:




So with the D's separated, we get the total possibilities minus the possibility of the two D's together:




You are doing your HSC in 2015 so I'm guessing you are a future 3-unit student?

 

[carolinevivi]

Yep, I'm doing it next year

 

[carolinevivi]

Thank you for the help

 

[anomalousdecay]

Thank you for the help

No problem. Probability wasn't my strongest point so just get someone else to check and to help you out in person (which may be more helpful).

Oh and welcome to Bored of Studies.

 

[braintic]

(1) Neither answer offered is correct.
The first read 40000 as 400000 so didn't take cases for 5 and 6 digits.
The second didn't read the 'no repetition' and didn't realise there are 6 digits on offer so also didn't take cases.

6 digits:
. Five-sixths of all possible numbers = (5/6) times 6! = 600

5 digits:
. All possible numbers starting with 4: 5P4 = 120
. AND
. Four-fifths of all possible numbers starting 5, 7 or 9: (4/5) times 3 times 5P4 = 288

Total: 1008


(3) Neither answer correct. The first thought the couple had to be together, but that is not stated. Also didn't find the probability.

Three groups: 2 adults and a single group of 5 children - 3! arrangements
Then arrange the 5 children within their group: 5! ways

So (3! times 5!) / 7! = 1/7

ALTERNATIVELY

Pick the seats for the couple: 7C2=21 possibilities (ignoring order)

Only 3 possibilities where the children are all together:
. Adults in first (left) two seats
. Adults in last (right) two seats
. Adults on either end

3/21 = 1/7

 

[anomalousdecay]

(1) Neither answer offered is correct.
The first read 40000 as 400000 so didn't take cases for 5 and 6 digits.
The second didn't read the 'no repetition' and didn't realise there are 6 digits on offer so also didn't take cases.

6 digits:
. Five-sixths of all possible numbers = (5/6) times 6! = 600

5 digits:
. All possible numbers starting with 4: 5P4 = 120
. AND
. Four-fifths of all possible numbers starting 5, 7 or 9: (4/5) times 3 times 5P4 = 288

Total: 1008


(3) Neither answer correct. The first thought the couple had to be together, but that is not stated. Also didn't find the probability.

Three groups: 2 adults and a single group of 5 children - 3! arrangements
Then arrange the 5 children within their group: 5! ways

So (3! times 5!) / 7! = 1/7

ALTERNATIVELY

Pick the seats for the couple: 7C2=21 possibilities (ignoring order)

Only 3 possibilities where the children are all together:
. Adults in first (left) two seats
. Adults in last (right) two seats
. Adults on either end

3/21 = 1/7

Yeah this is probably right haha.