Some combinations and Binomial questions (1 Viewer)

[GaganDeep]

22) IN how many ways can 3 boys and 2 girls be arranged in a row if a selection is made from 5 boys and 4girls?in how many of these arrangements does a boy occupy the middle position?
first part i got.
5 4
C * C * 5!
3 2
I don't know how to do the second part

1)Five cards are drawn at random from a pack of 52 playing ards. what is the probability that they are all from the same suit?
I got 13C5
--
52C5
But that's wrong?

7) Eigth people are to be divided into two groups. What is the probaility that there will be 4 in each group?

8)The letters of the word promise are arranged in a row. What is the probability that there are 3 letters between p and r, kinda of confused on this one

i am thinking 3!*3!*2!/7!
BUt that is wrong lol.

11) The number plate of a motor car contain 3 letters of the alphabet followed by 3 numerals. How many such number plates can be made? What proportion of these would contain 3 letters the same and 3 numerals the same?

22)Urn A containes 6 white and 4 black balls. Urb B contains 2 white and 2 black balles. From urn A two balles are selected at random and placed in Urn B. From Urn B two balls are then selected at random. What is the probability that exactly one of these two balls is white?

26) Four girls and four boys arrange themselves at randim in a row. What is the probability that the girls and the boys occupy alternate positions?
Isn;t it just
4!*4!/7!?


23) From a set of 10 cards numbered 1 to 10, two cards are drawn without replacement. What is the probability that
(a) both numbers are even
b)one is even and the other is odd
c)the sum of the two numbers is 12
d)both numbers are even and the sum of the two numbers is 12
27)Six people of whom A and B are two, arrange themselves at random in a row. What is the probability that there are atlest three people between A and B

SOme binonail questions

29) Simplify(Sqroot of 3 + 1)^6 + (sq root of 3 -1)^6
30)(sq root of (x-1) +1)^5 - (sqroot of (x-1) -1)^5
i don't want to expand and find out, is there a easier way? or another way
32) express as a binomial and evaluate, 0.998^20
=(1-0.002)^20
Do you need to expand to find out?
How do you do it? lol
Thanks i'm dumb need alot of help.

Also when find the greatest term and if your k values is an integer why do you need to check the term before it aswell.

Thanks in advance.

 

[Riviet]

1)Five cards are drawn at random from a pack of 52 playing ards. what is the probability that they are all from the same suit?

11) The number plate of a motor car contain 3 letters of the alphabet followed by 3 numerals. How many such number plates can be made? What proportion of these would contain 3 letters the same and 3 numerals the same?

1) P(5 same suit)=12/51x11/50x10/49x9/48 = 33/16660 because the first card picked can be of any suit but the ones after need to be the same.

11) Total number of number plates=26³.10³=17576000

Total proportion of number plates with 3 letters and 3 numerals the same=

(26x10)/17576000 = 1/67600.

I checked the answers at the bob and they're right.

 

[GaganDeep]

Oh k cool thanks. Pplease post when you have done the rest.

 

[insert-username]

29) Simplify(Sqroot of 3 + 1)^6 + (sq root of 3 -1)^6
30)(sq root of (x-1) +1)^5 - (sqroot of (x-1) -1)^5
i don't want to expand and find out, is there a easier way? or another way
32) express as a binomial and evaluate, 0.998^20
=(1-0.002)^20
Do you need to expand to find out?
How do you do it? lol

You don't need to expand it out. You need to use the binomial theorem, which relates to the Combination function. I haven't learnt this yet, so I can't help you more on this point.

26) Four girls and four boys arrange themselves at randim in a row. What is the probability that the girls and the boys occupy alternate positions?

I think this is right:

Number of possible rows = 8! = 40320

P(alternate) = 2 x (4C1 * 4C1 * 3C1 * 3C1 * 2C1 *2C1 * 1C1 * 1C1)/40320 [i.e. begin with any girl, any boy, etc, and the 2x is for any boy, any girl, etc)

= 1152/40320

= 1/35

23) From a set of 10 cards numbered 1 to 10, two cards are drawn without replacement. What is the probability that
(a) both numbers are even
b)one is even and the other is odd
c)the sum of the two numbers is 12
d)both numbers are even and the sum of the two numbers is 12

(a) 1/2 * 4/9 = 4/18 (If the first is even, there will be 4 even cards and 5 odd cards left)

(b) 1/2 * 5/9 = 5/18 (If the first is even/odd, there will be 5 odd/even cards left)

(c) 9/10 * 4/9 = 6/15 (You need one of 2, 3 ,4, 5, 7, 8, 9, 10 first, and the corresponding one second)

(d) 2/5 * 2/9 = 4/45 (You need one of 2, 4, 8, or 10 first, and the corresponding one second)


I_F

 

[Riviet]

8)The letters of the word promise are arranged in a row. What is the probability that there are 3 letters between p and r, kinda of confused on this one

There are three ways to arrange the letters of promise in a row where p is before r, ie pomirse, opmisre and ompiser, ignoring the other 5 letters. Now out of these 3 combos, you can rearrange the p and r, hence there are 2x3 ways. Now with the other 5 letters, there are 5! ways. Multiply 2x3x5! to get 720. Now the total number of arrangements without restriction= 7!

.: probability of three letters between p and r= 720/7! = 720/5040 = 1/7.