Help me with these problems...

Prakash Panneer · 2006-10-24 06:08:34

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.

Thanks in Advance

pi man · 2006-10-24 07:58:53

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%.

pi man · 2006-10-24 08:18:16

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)

pi man · 2006-10-24 08:27:16

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)

John E. Franklin · 2006-10-24 08:30:08

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

pi man · 2006-10-24 08:32:55

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%

pi man · 2006-10-24 08:41:56

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%

pi man · 2006-10-24 08:49:54

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.

pi man · 2006-10-24 09:01:05

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%

pi man · 2006-10-24 09:04:30

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

Math Is Fun Forum

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

Help me with these problems...

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

Re: Help me with these problems...

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

Re: Help me with these problems...

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

Re: Help me with these problems...

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

Re: Help me with these problems...

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

Re: Help me with these problems...

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

Re: Help me with these problems...

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

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

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

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