S1 and S2 DOUBTS HERE!!!!

[perish007]

this question is of statistics S1

Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the number is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.

[Tohru Kyo Sohma]

this question is of statistics S1

Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the digits is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.

im not sure but is it like this
(a)6!/3!
(b)i didnt get the qns
i dont remember much but if u have the answer u can check with it

[astarmathsandphysics]

Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the number is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.

Must have 2 4's, or 2 5's or both.
If 2 4's there are 6!/2! different numbers and same for two 5's.
If 2 4's and two 5's then there are 7!/2!2! ways for each choice of the other two numbers. There are three ways the other two can be chosen so altogether 2*6!/2! +3*6!/2! ways altogether.

a) if the number is odd then the last digit is 3, 5 or 7.
If the last digit is 3, then the other 5 numbers can be chosen from 4,4,5,5,6,7
Following the logic above there are 2*5!/2! +2*5!/2! ways of choosing these five. The same result is is 7 is the last number.
If 5 is the last number, and the other 5 numbers contain one 4, there are 5! ways and 5!/2! ways if two 4's.
Hence preb of being odd is (2(2*5!/2! +2*5!/2!)+5!+5!/2!)/2*6!/2! +3*6!/2!)
b)Two 5's so one 4. There are 6!/2! ways this can be down. Prob = 6!/2!/(Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the number is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.

Must have 2 4's, or 2 5's or both.
If 2 4's there are 6!/2! different numbers and same for two 5's.
If 2 4's and two 5's then there are 6!/2!2! ways for each choice of the other two numbers. There are three ways the other two can be chosen so altogether 2*6!/2! +3*6!/2! ways altogether.

a) if the number is odd then the last digit is 3, 5 or 7.
If the last digit is 3, then the other 5 numbers can be chosen from 4,4,5,5,6,7
Following the logic above there are 2*5!/2! +2*5!/2! ways of choosing these five. The same result is is 7 is the last number.
If 5 is the last number, and the other 5 numbers contain one 4, there are 5! ways and 5!/2! ways if two 4's.
Hence preb of being odd is (2(2*5!/2! +2*5!/2!)+5!+5!/2!)/2*6!/2! +3*6!/2!)
b)Two 5's so one 4. There are 6!/2! ways this can be done so 6!/2!/(2*6!/2! +3*6!/2!)

[ashwinkandel]

Eight People sit in a minibus: four on the sunny side and four on the shady side. If two people want to sit on opposite sides to each other and another two people want to sit on the shady sides, in how many ways can this be done?

[astarmathsandphysics]

here

[ashwinkandel]

Thanks for clarifying this.. but you have written 6! instead of 4!

[astarmathsandphysics]

you are right. Should be 4!

[ashwinkandel]

Disco Lights are arranged in a vertical line. How many different arrangements can be made from two green, three blue and four red lights if at least eight lights are used?

[astarmathsandphysics]

eight lights used
1 red three blue four red 1!*3!*4!
2 red two blue four red 2!*2!*4!
2 red three blue three red 2!*3!*3!
nine lights used 2!*3!*4!
add all these up

[ashwinkandel]

Three letters are selected at random from the word SCHOOL. Find the probability that the selection
i. does not contain the letter O,
ii. contains both the letters O.

[iluvme]

i. Is it 4/6?

Doing probability maybe after 2 years.

[ashwinkandel]

For both problems, answers are given as 1/5  i took out 2/7 for both.

[iluvme]

For both problems, answers are given as 1/5  i took out 2/7 for both.

Sorry maybe someone else would answer.

But why 2/7?

[ashwinkandel]

No. of selections with no O's=4c3=4
No. of selections with one O= 4C2=6
No. of selections with two O's=4C1=4
Total Selections=6+4+4=14
P(Does not contain letter O)=4/14=2/7
P(Contains both the letters O)=4/14=2/7
This is how i did but the answers given were 1/5. I think my solution is right but want to confirm it.

[~ Miss Relina ~]

attached Q4 part (b0
and Q5 plz