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  • Jul 31, 2012, 04:28 AM
    ccny_miah
    statistics help free online
    republican democrat independent
    male 46 39 1
    female 5 9 0

    1. randomly select one senator, what is the probability of getting a republican, given that a male was selected?

    2. randomly select one senator, what is the probability of getting a male, given that a republican was selected?

    3. randomly select 1 senator, what is the probability of getting a female, given that an independent was selected?

    4. randomly select 1 senator, what is the probability of getting a democrat or independent, given that a male was selected?

    5. randomly select 2 senators, what is the probability of getting a democrat female and a republican male?
  • Jul 31, 2012, 04:34 AM
    Curlyben
    So what do YOU think ?
  • Jul 31, 2012, 04:42 AM
    ccny_miah
    1. 46/86?
  • Jul 31, 2012, 05:13 AM
    ccny_miah
    Quote:

    Originally Posted by Curlyben View Post
    So what do YOU think ?

    Is it correct?
  • Jul 31, 2012, 05:47 AM
    ebaines
    Quote:

    Originally Posted by ccny_miah View Post
    1. 46/86?

    Correct - there are 46 republicans out of 86 males. Of course you can simplify that fraction.

    What answers did you come up with for the other questions?
  • Jul 31, 2012, 05:57 AM
    ccny_miah
    2. 46/86?
  • Jul 31, 2012, 06:01 AM
    ebaines
    Quote:

    Originally Posted by ccny_miah View Post
    2. 46/86?

    Correct numerator. But the denominator should be equal to the total number of republicans - what number is that?
  • Jul 31, 2012, 06:27 AM
    ccny_miah
    Quote:

    Originally Posted by ebaines View Post
    Correct numerator. But the denominator should be equal to the total number of republicans - what number is that?

    I meant, 46/51.
  • Jul 31, 2012, 06:32 AM
    ccny_miah
    3. 0/1?
  • Jul 31, 2012, 08:25 AM
    ebaines
    Yes - answers for 2 and 3 are both correct. It seems you have the technique down nicely. Looking forward to seeing what you get for number 5.
  • Jul 31, 2012, 09:34 AM
    ccny_miah
    Quote:

    Originally Posted by ebaines View Post
    Yes - answers for 2 and 3 are both correct. It seems you have the technique down nicely. Looking forward to seeing what you get for number 5.

    4. 39+1 over 46+39+1
    40/86?
  • Jul 31, 2012, 09:39 AM
    ccny_miah
    And I don't get question 5. please help!
  • Jul 31, 2012, 09:59 AM
    ebaines
    Quote:

    Originally Posted by ccny_miah View Post
    4. 39+1 over 46+39+1
    40/86?

    Good.
  • Jul 31, 2012, 10:11 AM
    ebaines
    Quote:

    Originally Posted by ccny_miah View Post
    and i dont get question 5. please help!

    To get a dem female (DF) and a rep male (RM) means you can either (a) select a dem female followed by a rep male, or (b) select a rep male followed by a dem female. The probabiliy of (a) is



    Note that the numerator of each fraction is the number of senators of appropriate persuassion, ane the denominator is the total number of senators to pick from.

    Using this same technique you can calculate the probability of (b) - what do you get? Finally add to get the total probability.
  • Jul 31, 2012, 10:51 AM
    ccny_miah
    9x46/100x99= 23/550 therefore, (23/550)+(23/550)= 46/550. We can reduce to 23/275. The final answer should be 23/275. Correct?
  • Jul 31, 2012, 11:03 AM
    ebaines
    Quote:

    Originally Posted by ccny_miah View Post
    9x46/100x99= 23/550 therefore, (23/550)+(23/550)= 46/550. we can reduce to 23/275. the final answer should be 23/275. Correct?

    Yes - correct!
  • Jul 31, 2012, 11:15 AM
    ccny_miah
    Thank you very much.

    Question: the starting 5 players for the NY Knicks team have agreed to make charity appearance. If you must send 3 player to United Way event and the other 2 to Heart Fund event, how many different ways can you make the assignments?
  • Jul 31, 2012, 11:22 AM
    ccny_miah
    5! / (2)!(3)!= 10? Used the nCr formula.
  • Jul 31, 2012, 11:27 AM
    ebaines
    Quote:

    Originally Posted by ccny_miah View Post
    5! / (2)!(3)!= 10? used the nCr formula.

    Yep.
  • Jul 31, 2012, 11:39 AM
    ccny_miah
    winning power ball lottery requires you to select 5 numbers between 1 and 53 inclusive and, in a separate drawing, you must also select the correct single number between 1 and 42 inclusive. Find the probability of winning the jackpot?
  • Jul 31, 2012, 11:40 AM
    ccny_miah
    I don't know where to start.
  • Jul 31, 2012, 11:47 AM
    ebaines
    The numerator is the number of ways that you can select the winning 5 numbers times the number of ways you can select the correct extra number. The denominator is the total number of possible ways that 5 balls can be drawn from 53, times the number of ways that the one additional ball can be drawn from 42.
  • Jul 31, 2012, 11:59 AM
    ccny_miah
    Quote:

    Originally Posted by ebaines View Post
    The numerator is the number fo ways that you can select the winning 5 numbers times the number if ways you can select the correct extra number. The denominator is the total number of possible ways that 5 balls can be drawn from 53, times the number of ways that the one additiobnal ball can be drawn from 42.

    Can you please show me... I really confused on this.
  • Jul 31, 2012, 12:44 PM
    ebaines
    I'll give you a different example that illustrates the thought process. What's the probability of dealing a royal straight spade flush from a deck of 52 cards? A royal straight spade flush consists of the ace of spades, king of spades, queen of spades, jack of spades, and 10 of spades. It doesn't matter what order the five cards are dealt in. The number of ways those five cards can be dealt is 5! The total number of possible ways to deal five cards is 52 x 51 x 50 x 49 x 48 = 52P5 (permutations of 52 cards 5 at a time). Hence the odds of dealing a royal straight flush is 5!/(52P5) = 0.0000385%.
  • Aug 1, 2012, 06:35 AM
    ccny_miah
    Quote:

    Originally Posted by ebaines View Post
    The numerator is the number of ways that you can select the winning 5 numbers times the number of ways you can select the correct extra number. The denominator is the total number of possible ways that 5 balls can be drawn from 53, times the number of ways that the one additional ball can be drawn from 42.

    53C5 x 42C1

    53!x42/(53-5)!(5!)

    (53x52x51x50x49)(42)/5x4x3x2x1
    = 120,526,770?

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