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## #1 2006-10-24 06:08:34

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

### Help me with these problems...

1)A bag contains 5 red , 4 green and 3 black balls . Three balls are drawn out of it at random, find the probability  of drawing exactly 2 red balls.

2)There are 4 envelopes corresponding to 4 letters. If the letters are placed in the envelopes at random, what is the probability that all the letters are not placed in the right envelopes.

3)From a pack of 52 playing cards, three cards are drawn at random . Find the probability of drawing a king, a queen and a jack.

4)If a leap year is selected at random, what is  the chance that it will contain 53 Tuesdays?

5)What are the odds in favour of throwing at least 8 in a single throw with two dice?

6)A coin is tossed successively 3 times . Find the probability of getting exactly one head or two heads.
7)6 boys and 6 girls sit in a row randomly , find the probability that all the 6 girls sit together.
8) A boy has 6 pockets . In how many ways can he put  5 coins in his pockets?

9)In how many ways can 5 letters be posted in 7 letter boxes in a town?  If all the letters are not posted in the same  box , find the corresponding number of ways of posting .

10)Find the number of 3-digit numbers that can be formed with the digits  1,3,5,7,9
when  a digit may repeated any number of  times.

Letter, number, arts and science
of living kinds, both are the eyes.

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## #2 2006-10-24 07:58:53

pi man
Member
Registered: 2006-07-06
Posts: 251

### Re: Help me with these problems...

1.   There are 12 marbles in all.   You want to randomy pick 3 of them.   So there are (12 choose 3) ways to do that:

You want to know what the odds of drawing 2 red balls plus either a green or black ball.   There are 10 different combinations of 2 red balls (5 choose 2) and there are 7 different greeen or black balls (7 choose 1) to choose from.   So that's 10*7 different ways out of the possible 220 ways.   70/220 ~= 32%.

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## #3 2006-10-24 08:18:16

pi man
Member
Registered: 2006-07-06
Posts: 251

### Re: Help me with these problems...

2.  There are 24 different ways (4!)  to put the 4 letters (a, b, c and d)  in the 4 envelopes (A, B, C and D).

There is only 1 chance out of the 24 combinations of getting all 4 letters in the right envelope.
How many ways of getting exactly 3 of the 4 letters right?   Zero.   If three of them are right, the 4th would have to be right also.

How many ways of getting exactly 2 of the letters right?  Lets say we get the first two correct (Aa, Bb).   That means c has to go in D and d has to go in C:  Aa, Bb, Cd, Dc.    So there's only 1 way to get only letters a and b in the correct envelope.   There's also other ways to get 2 letters right - a and c, a and d, b and c, b and d, and c and d (4 choose 2).  So there are 6 different ways to get exactly 2 letters in the correct envelope.

How many ways of getting exactly 1 letter right?   Let's say D is correct.    The possible combinations for the first 3 are ABC, ACB, BAC, BCA, CAB and CBA.   Only 2 of these combinations have the none of the letters in the right envelope - BCA and CAB.   So there are 2 ways to get only letter D in the correct envelope.   Similarily there are 2 ways each to get only A, B or C in the correct envelope.   That gives a total of 8 ways to get just one right.

How many ways of getting exactly 0 letters right?   THere are 24 combinations total.   1 way to get 4 right, 6 ways to get 2 right, 8 ways to get one right.   That leaves 9 ways to get 0 right (24 - 1 - 6 - 8)

Last edited by pi man (2006-10-24 08:28:41)

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## #4 2006-10-24 08:27:16

pi man
Member
Registered: 2006-07-06
Posts: 251

### Re: Help me with these problems...

3.  There are 52 choose 3 ways of picking 3 cards from a deck of 52.

You want to choose 1 of the four possible jacks, one of the four possible queens and one of four possible kings (4 choose 1 for each).

There are 64 different ways to choose 1 jack, 1 queen and 1 king.   That's out of 22100 possibilites, so the probabily is 64 / 22100 or 2.896%

Last edited by pi man (2006-10-24 08:28:55)

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## #5 2006-10-24 08:30:08

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: Help me with these problems...

#4  Probability is two-sevenths because there are 52 weeks and 2 days in a leap year.

igloo myrtilles fourmis

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## #6 2006-10-24 08:32:55

pi man
Member
Registered: 2006-07-06
Posts: 251

### Re: Help me with these problems...

4.   There are 366 days in a leap year.    That's 52 full weeks plus 2 extra days.    If the year starts on a Sunday, there's going to be 52 full weeks of Sunday through Saturday, plus an extra Sunday and a Monday.   Likewise, if it beings on a Monday, there's going to be 53 Monday's and 53 Tuesday's.    There's going to be 53 Tuesdays whenever the year starts on a Monday or a Tuesday.   So that's 2 of the possible 7 days the year could start  on.   2 / 7 = 28.57%

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## #7 2006-10-24 08:41:56

pi man
Member
Registered: 2006-07-06
Posts: 251

### Re: Help me with these problems...

5.   Ways of throwing a particular total when throwing 2 dices:

2   - 1 (1,1)
3   - 2 (1,2), (2,1)
4   - 3 (1,3), (2,2), (3,1)
5   - 4 (1,4) (2,3), (3,2), (4,1)
6   - 5 (1,5), (2,4), (3,3), (4,2), (5,1)
7   - 6 (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
8   - 5 (2,6), (3,5), (4,4), (5,3), (6,2)
9   - 4 (3,6), (4,5), (5,4), (6,3)
10 - 3 (4,6), (5,5), (6,4)
11 - 2 (5,6), (6,5)
12 - 1 (6,6)

So there are 15 (5+4+3+2+1) ways of throwing a total of 8 or better.  8 of 36 is 22.22%

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## #8 2006-10-24 08:49:54

pi man
Member
Registered: 2006-07-06
Posts: 251

### Re: Help me with these problems...

6.   Find the probability of getting 1 or 2 heads exactly when flipping a coin 3 times.   It's easier to consider the opposite:  what's the odds of getting 0 or 3 heads exactly?   Theres only 1 way each to get those two results (TTT, HHH).    There's 8 differents ways total (2*2*2) so that leaves 6 of 8 chances (75%) of getting 1 or 2 heads.

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## #9 2006-10-24 09:01:05

pi man
Member
Registered: 2006-07-06
Posts: 251

### Re: Help me with these problems...

There are 12! ways to arrange the 12 kids.   That's 479,001,600. diffent ways.

For all of the girls to be sitting in a row, they could be occupying seats 1-6, 2-7, 3-8, 4-9, 5-10, 6-11 or 7-12.  That's 7 different grops of seats that could be occupying.    Within each of those groups, they could be arranged 6! different ways (720) for a total of 7*720=5040 different ways.   5040 ways out of 479001600 = .001052%

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## #10 2006-10-24 09:04:30

pi man
Member
Registered: 2006-07-06
Posts: 251

### Re: Help me with these problems...

10.  Since any digit can be repeated, there are 5 possibilities for the first digit, 5 for the second, and 5 for the third.  Or 5*5*5=125 different numbers

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