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anonymous
 5 years ago
any one good with pobability?
anonymous
 5 years ago
any one good with pobability?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0maybe..but then again.....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0What is the probablitity of typing the word probability with no mistakes 5 times in a row?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i have an assignment im not quite understanding.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cant read your mind...spit it out :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0her assignment is to find out what the probability is that you will randomly type the correct answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A 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
 5 years ago
Best ResponseYou've already chosen the best response.0Suppose 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
 5 years ago
Best ResponseYou've already chosen the best response.0there 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
 5 years ago
Best ResponseYou've already chosen the best response.0the probability of a red face card is 6 out of 52; (6/62) x100 for a percent

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is it a red jack? or a jack or a red card?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04jack 24red cards  *  seems right. what you think? 52 52

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or it could mean you have a 26/52 chance of drawing either a red card or a jack.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would it be 6/52 or 62?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0What does P(ab) mean: a or b right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im guessing a red jack

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ya i believe so? where are you seeing that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0just recalling that in programing that "" bar means or; and after checking up on it, I was right.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0P(ab) = P(a) + P(b)  P(A & B).

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Pjack = 4/52 Pred = 26/52 Pred jack = 2/52 (4+26)  2  = 28/52 52

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0P(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+434  = 43/52 52

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, just going over what you are doing so I can do it on the next problem

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this 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
 5 years ago
Best ResponseYou've already chosen the best response.0Suppose 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
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you for helping me and breakin it down to steps.
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