anonymous
  • anonymous
any one good with pobability?
Mathematics
schrodinger
  • schrodinger
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amistre64
  • amistre64
maybe..but then again.....
amistre64
  • amistre64
What is the probablitity of typing the word probability with no mistakes 5 times in a row?
anonymous
  • anonymous
well i have an assignment im not quite understanding.

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amistre64
  • amistre64
cant read your mind...spit it out :)
anonymous
  • anonymous
her assignment is to find out what the probability is that you will randomly type the correct answer
anonymous
  • anonymous
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) =
anonymous
  • anonymous
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.
amistre64
  • amistre64
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
amistre64
  • amistre64
the probability of a red face card is 6 out of 52; (6/62) x100 for a percent
amistre64
  • amistre64
is it a red jack? or a jack or a red card?
amistre64
  • amistre64
4jack 24red cards ----- * ------------ seems right. what you think? 52 52
amistre64
  • amistre64
or it could mean you have a 26/52 chance of drawing either a red card or a jack.
anonymous
  • anonymous
would it be 6/52 or 62?
amistre64
  • amistre64
What does P(a|b) mean: a or b right?
anonymous
  • anonymous
im guessing a red jack
anonymous
  • anonymous
ya i believe so? where are you seeing that
amistre64
  • amistre64
just recalling that in programing that "|" bar means or; and after checking up on it, I was right.
amistre64
  • amistre64
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.
amistre64
  • amistre64
P(a|b) = P(a) + P(b) - P(A & B).
amistre64
  • amistre64
Pjack = 4/52 Pred = 26/52 Pred jack = 2/52 (4+26) - 2 --------- = 28/52 52
amistre64
  • amistre64
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
amistre64
  • amistre64
make sense?
anonymous
  • anonymous
yes, just going over what you are doing so I can do it on the next problem
amistre64
  • amistre64
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
amistre64
  • amistre64
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"
anonymous
  • anonymous
Thank you for helping me and breakin it down to steps.

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