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Of course...I would be procrastinating on data management homework -.-...

Anyways would anyone care to explain how to solve the following two questions?

*hoping tht the questions are hard and its not one of those ones where when you find out the answer u hit urself in the forehead cuz it was actually simple*

1)Suppose the two joker cards are left in a standard deck of cards. One of the jokers is red and the other black. A single card is drawn from the deck of 54 cards, returned, and then a second card is drawn. Determine the probability of drawing the red joker or a red ace on either draw.

*I know how to get the answer...but I do not understand why you need to use the complementary event equation

2)A health and safety committee is to be selected from all the people who work at a local factory. The committeeis to consist of four members selected randomly from a list of ten names submitted by the shop leader. The list has the names of five union members and five workers who are not union members. What is the probability that three of the four committee members are union members?

Please and thank you to anyone willing to try ^-^

Anyways would anyone care to explain how to solve the following two questions?

*hoping tht the questions are hard and its not one of those ones where when you find out the answer u hit urself in the forehead cuz it was actually simple*

1)Suppose the two joker cards are left in a standard deck of cards. One of the jokers is red and the other black. A single card is drawn from the deck of 54 cards, returned, and then a second card is drawn. Determine the probability of drawing the red joker or a red ace on either draw.

*I know how to get the answer...but I do not understand why you need to use the complementary event equation

2)A health and safety committee is to be selected from all the people who work at a local factory. The committeeis to consist of four members selected randomly from a list of ten names submitted by the shop leader. The list has the names of five union members and five workers who are not union members. What is the probability that three of the four committee members are union members?

Please and thank you to anyone willing to try ^-^

9
replies

1) I think this question refers to the probability of drawing a red ace or a red joker at least once in two draws (with replacement), correct me if I'm wrong since the wording is somewhat ambiguous.

At least one of your draws has a red ace or a red joker. It's possible that you draw a red joker and a red ace both times.

To avoid double counting, after you calculate the probability of drawing one of the two cards on either draw, you should subtract the probability of drawing one of the two BOTH times. I'm not sure if the complementary event equation has to be involved, though. In any given draw, your chance of drawing one of the cards is 2/54 = 1/27.

Anyway, I came up with two approaches:

Approach 1: Union (Summing the two draws, subtracting the intersection)

P(Event) = 1/27 (first draw) + 1/27 (second draw) - (1/27)^2 (drawing both times)

= 2/27 - 1/729

= 54/729 - 1/729

= 53/729 = about 7.27%

Approach 2: Exponential distribution (waiting time, until you draw one of the desired cards)

P(Event) = 1/27 (success after one draw) + (1/27)(26/27) (success after one failed draw)

= 1/27 + 26/729

= 27/729 + 26/729

= 53/729 = about 7.27%

2) In this example, you choose 3 people (out of 5) from the union and 1 person (out of 5)not from the union.

So, N(Event)= (5C3)*(5C1) = 5!/3!2! * 5!/1!4! = 10*5 = 50

There are 50 possible ways to choose the four committee members given the conditions. Let's find the sample set now, where you are simply picking 4 out of 10 in random order.

N(S)= 10C4 = 10!/6!4! = 210

Computing the probability of event E, P(Event)= N(Event)/N(S) = 50/210 = 5/21 or about 23.8%.

At least one of your draws has a red ace or a red joker. It's possible that you draw a red joker and a red ace both times.

To avoid double counting, after you calculate the probability of drawing one of the two cards on either draw, you should subtract the probability of drawing one of the two BOTH times. I'm not sure if the complementary event equation has to be involved, though. In any given draw, your chance of drawing one of the cards is 2/54 = 1/27.

Anyway, I came up with two approaches:

Approach 1: Union (Summing the two draws, subtracting the intersection)

P(Event) = 1/27 (first draw) + 1/27 (second draw) - (1/27)^2 (drawing both times)

= 2/27 - 1/729

= 54/729 - 1/729

= 53/729 = about 7.27%

Approach 2: Exponential distribution (waiting time, until you draw one of the desired cards)

P(Event) = 1/27 (success after one draw) + (1/27)(26/27) (success after one failed draw)

= 1/27 + 26/729

= 27/729 + 26/729

= 53/729 = about 7.27%

2) In this example, you choose 3 people (out of 5) from the union and 1 person (out of 5)not from the union.

So, N(Event)= (5C3)*(5C1) = 5!/3!2! * 5!/1!4! = 10*5 = 50

There are 50 possible ways to choose the four committee members given the conditions. Let's find the sample set now, where you are simply picking 4 out of 10 in random order.

N(S)= 10C4 = 10!/6!4! = 210

Computing the probability of event E, P(Event)= N(Event)/N(S) = 50/210 = 5/21 or about 23.8%.

If my answers were not helpful (incorrect, misleading or confusing), I apologize in advance. I read your post this very, very late at night.

@superstar2011 wrote

If my answers were not helpful (incorrect, misleading or confusing), I apologize in advance. I read your post this very, very late at night.

Thank you for your answers. Unfortunately, the problem is tht the answers are not the same as the back of the textbook...even though there have been times that they are wrong...my teacher "assured" me that the answers for these particular questions are correct but again thank you. If you are at all curious to what the answers were they were: 35/324 for the first one so approximately 11% and then 5/84 approximately being 6% for the second one.

@RheyDCD11 wrote

@superstar2011 wrote

If my answers were not helpful (incorrect, misleading or confusing), I apologize in advance. I read your post this very, very late at night.

Thank you for your answers. Unfortunately, the problem is tht the answers are not the same as the back of the textbook...even though there have been times that they are wrong...my teacher "assured" me that the answers for these particular questions are correct but again thank you. If you are at all curious to what the answers were they were: 35/324 for the first one so approximately 11% and then 5/84 approximately being 6% for the second one.

I guess that's the problem with doing this late at night. I'll repost my answers in case they are of any help.

1) What I forgot here is that there are TWO red aces in the deck; an ace of hearts and an ace of diamonds. So there are not two, but three cards in the deck in which you would like to draw. My methods, however, were correct.

At least one of your draws has a red ace or a red joker. It's possible that you draw a red joker and a red ace both times.

In any given draw, your chance of drawing one of the cards is 3/54 = 1/18.

Using the same approaches as I did before,

Approach 1: Union (Summing the two draws, subtracting the intersection)

P(Event) = 1/18 (first draw) + 1/18 (second draw) - (1/18)^2 (drawing both times)

= 1/9 - 1/324

= 36/324 - 1/324

= 35/324 = about 10.8%

Approach 2: Exponential distribution (waiting time, until you draw one of the desired cards)

P(Event) = 1/18 (success after one draw) + (1/18)(17/18) (success after one failed draw)

= 1/18 + 17/324

= 18/324 + 17/324

= 35/324 = about 10.8%

2) I think the problem with this question is that it wasn't specific, but rather ambiguous; apparently the order in which the committee members were picked mattered. (I don't know why, were they assigned different roles?)

Okay, so I will use the counting principle that I used before, except with permutations instead of combinations (now that order matters).

In this example, you choose 3 people (out of 5) from the union and 1 person (out of 5) not from the union.

So, N(Event)= (5P3)*(5P1) = 5!/2! * 5!/4! = 60*5 = 300

There are 300 possible ways to pick 3 union members and 1 non-union member in a particular order. Let's find the sample set now, where you are simply picking 4 members from a list of 10.

N(S)= 10P4 = 10!/6! = 5040

Computing the probability of event E, P(Event)= N(Event)/N(S) = 300/5040 = 5/84 (divide both numbers by 60) or about 5.95%.

@superstar2011 wrote

@RheyDCD11 wrote

@superstar2011 wrote

If my answers were not helpful (incorrect, misleading or confusing), I apologize in advance. I read your post this very, very late at night.

Thank you for your answers. Unfortunately, the problem is tht the answers are not the same as the back of the textbook...even though there have been times that they are wrong...my teacher "assured" me that the answers for these particular questions are correct but again thank you. If you are at all curious to what the answers were they were: 35/324 for the first one so approximately 11% and then 5/84 approximately being 6% for the second one.

I guess that's the problem with doing this late at night. I'll repost my answers in case they are of any help.

1) What I forgot here is that there are TWO red aces in the deck; an ace of hearts and an ace of diamonds. So there are not two, but three cards in the deck in which you would like to draw. My methods, however, were correct.

At least one of your draws has a red ace or a red joker. It's possible that you draw a red joker and a red ace both times.

In any given draw, your chance of drawing one of the cards is 3/54 = 1/18.

Using the same approaches as I did before,

Approach 1: Union (Summing the two draws, subtracting the intersection)

P(Event) = 1/18 (first draw) + 1/18 (second draw) - (1/18)^2 (drawing both times)

= 1/9 - 1/324

= 36/324 - 1/324

= 35/324 = about 10.8%

Approach 2: Exponential distribution (waiting time, until you draw one of the desired cards)

P(Event) = 1/18 (success after one draw) + (1/18)(17/18) (success after one failed draw)

= 1/18 + 17/324

= 36/324 + 17/324

= 35/324 = about 10.8%

2) I think the problem with this question is that it wasn't specific, but rather ambiguous; apparently the order in which the committee members were picked mattered. (I don't know why, were they assigned different roles?)

Okay, so I will use the counting principle that I used before, except with permutations instead of combinations (now that order matters).

In this example, you choose 3 people (out of 5) from the union and 1 person (out of 5) not from the union.

So, N(Event)= (5P3)*(5P1) = 5!/2! * 5!/4! = 60*5 = 300

There are 300 possible ways to pick 3 union members and 1 non-union member in a particular order. Let's find the sample set now, where you are simply picking 4 members from a list of 10.

N(S)= 10P4 = 10!/6! = 5040

Computing the probability of event E, P(Event)= N(Event)/N(S) = 300/5040 = 5/84 (divide both numbers by 60) or about 5.95%.

THANK YOU SO MUCH! the first one makes more sense to me than my teacher's initial approach...however its understandable why i did not get the first one since we have yet to learn combinations and we just learned permutations xD BUT THANK YOU A THOUSAND TIMES ^-^

You're very welcome. Ask questions anytime. Also, if you know the elements in Pascal's Triangle, combinations will make your life much easier by the time the organized counting unit is over. In combinations, unlike permutations, the order in which you pick does not matter.

Believe it or not, t(n,r) in Pascal's Triangle translates to nCr. For example, 5C3 = 10 because the 3rd element in row 5 of Pascal's Triangle is 10.

Believe it or not, t(n,r) in Pascal's Triangle translates to nCr. For example, 5C3 = 10 because the 3rd element in row 5 of Pascal's Triangle is 10.

@superstar2011 wrote

You're very welcome. Ask questions anytime. Also, if you know the elements in Pascal's Triangle, combinations will make your life much easier by the time the organized counting unit is over. In combinations, unlike permutations, the order in which you pick does not matter.

Believe it or not, t(n,r) in Pascal's Triangle translates to nCr. For example, 5C3 = 10 because the 3rd element in row 5 of Pascal's Triangle is 10.

haha can't wait dont worry there are probaly more questions in the future of my data management course -.-

Can someone help me with this question?? Please and thank you!!!

Before the invention of the telephone, Sam Morse developed an efficient system fro sending messages as a series of dots and dashes (short or long pulses). International code, a modified version of morse code is still widely used.

b) How many pulses would be necessary to represent the 72 letters of the Cambodian alphabet using a system like Morse code.

I have a quiz tomorrow and I'm so nervous! I really need to do well to bring my mark up D:

I've googled it and seen all the yahoo answers for it but I feel I still don't understand it very well. :/

Before the invention of the telephone, Sam Morse developed an efficient system fro sending messages as a series of dots and dashes (short or long pulses). International code, a modified version of morse code is still widely used.

b) How many pulses would be necessary to represent the 72 letters of the Cambodian alphabet using a system like Morse code.

I have a quiz tomorrow and I'm so nervous! I really need to do well to bring my mark up D:

I've googled it and seen all the yahoo answers for it but I feel I still don't understand it very well. :/

@terryaki26 wrote

Can someone help me with this question?? Please and thank you!!!

Before the invention of the telephone, Sam Morse developed an efficient system fro sending messages as a series of dots and dashes (short or long pulses). International code, a modified version of morse code is still widely used.

b) How many pulses would be necessary to represent the 72 letters of the Cambodian alphabet using a system like Morse code.

I have a quiz tomorrow and I'm so nervous! I really need to do well to bring my mark up D:

I've googled it and seen all the yahoo answers for it but I feel I still don't understand it very well. :/

Dots and dashes are two possibilities for a pulse. To have 72 different unique codes out of two symbols, you would have to have 'n' pulses, where 2^n = 72. 'n' must be a discrete number and must be rounded up.

2^n = 72

ln (2^n) = ln 72

n*ln 2 = ln 72

n = ln(72)/ln(2)

Since ln(72)/ln(2) is approximately equal to 6.17, you would have to have at least 7 pulses to uniquely represent each letter.