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anonymous

  • 5 years ago

any one good with pobability?

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  1. amistre64
    • 5 years ago
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    maybe..but then again.....

  2. amistre64
    • 5 years ago
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    What is the probablitity of typing the word probability with no mistakes 5 times in a row?

  3. anonymous
    • 5 years ago
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    well i have an assignment im not quite understanding.

  4. amistre64
    • 5 years ago
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    cant read your mind...spit it out :)

  5. anonymous
    • 5 years ago
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    her assignment is to find out what the probability is that you will randomly type the correct answer

  6. anonymous
    • 5 years ago
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    A single card is drawn from a standard deck of cards. Find the following probabilities. A face card is a jack, queen, or king. P(jack | red) = P(three | not a face card) =

  7. anonymous
    • 5 years ago
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    Suppose a pair of dice are rolled. Consider the sum of the numbers on the top of the dice and find the probabilities. (a) 8, given that exactly one die came up 5. (b) 4, given that exactly one die came up 3. (c) 6, given that exactly one die came up 6.

  8. amistre64
    • 5 years ago
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    there are 52 cards to a deck; with 4 suits and 3 face cards per suit. 12 face cards in the whole deck; 6 are red, and 6 are black

  9. amistre64
    • 5 years ago
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    the probability of a red face card is 6 out of 52; (6/62) x100 for a percent

  10. amistre64
    • 5 years ago
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    is it a red jack? or a jack or a red card?

  11. amistre64
    • 5 years ago
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    4jack 24red cards ----- * ------------ seems right. what you think? 52 52

  12. amistre64
    • 5 years ago
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    or it could mean you have a 26/52 chance of drawing either a red card or a jack.

  13. anonymous
    • 5 years ago
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    would it be 6/52 or 62?

  14. amistre64
    • 5 years ago
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    What does P(a|b) mean: a or b right?

  15. anonymous
    • 5 years ago
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    im guessing a red jack

  16. anonymous
    • 5 years ago
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    ya i believe so? where are you seeing that

  17. amistre64
    • 5 years ago
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    just recalling that in programing that "|" bar means or; and after checking up on it, I was right.

  18. amistre64
    • 5 years ago
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    but the chances of picking a jack versus or a red card has some issue to it, since a jack can also be a red card.

  19. amistre64
    • 5 years ago
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    P(a|b) = P(a) + P(b) - P(A & B).

  20. amistre64
    • 5 years ago
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    Pjack = 4/52 Pred = 26/52 Pred jack = 2/52 (4+26) - 2 --------- = 28/52 52

  21. amistre64
    • 5 years ago
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    P(three | not a face card) since 3 is not a face card, both of these occur at the same time. Pthree = 4/52 Pnot face = 43/52 4+43-4 ------- = 43/52 52

  22. amistre64
    • 5 years ago
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    make sense?

  23. anonymous
    • 5 years ago
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    yes, just going over what you are doing so I can do it on the next problem

  24. amistre64
    • 5 years ago
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    this dice one is not possible: (c) 6, given that exactly one die came up 6. the sum of the dice would at least have to be 7 if one is a 6. 6 + 1 = 7; so the sum being 6 between the two dice is impossible

  25. amistre64
    • 5 years ago
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    Suppose a pair of dice are rolled. Consider the sum of the numbers on the top of the dice and find the probabilities. (a) 8, given that exactly one die came up 5. Since we dont need to consider the other dice here, we should just simply figure the probability of getting a 3 on the other one. which is a 1/6 chance. (b) 4, given that exactly one die came up 3. same logic here; ther eis a 1/6 chance of the other die being a "1"

  26. anonymous
    • 5 years ago
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    Thank you for helping me and breakin it down to steps.

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