any one good with pobability?

- anonymous

any one good with pobability?

- schrodinger

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- amistre64

maybe..but then again.....

- amistre64

What is the probablitity of typing the word probability with no mistakes 5 times in a row?

- anonymous

well i have an assignment im not quite understanding.

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## More answers

- amistre64

cant read your mind...spit it out :)

- anonymous

her assignment is to find out what the probability is that you will randomly type the correct answer

- 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

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

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

the probability of a red face card is 6 out of 52; (6/62) x100 for a percent

- amistre64

is it a red jack? or a jack or a red card?

- amistre64

4jack 24red cards
----- * ------------ seems right. what you think?
52 52

- amistre64

or it could mean you have a 26/52 chance of drawing either a red card or a jack.

- anonymous

would it be 6/52 or 62?

- amistre64

What does P(a|b) mean: a or b right?

- anonymous

im guessing a red jack

- anonymous

ya i believe so? where are you seeing that

- amistre64

just recalling that in programing that "|" bar means or; and after checking up on it, I was right.

- 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

P(a|b) = P(a) + P(b) - P(A & B).

- amistre64

Pjack = 4/52
Pred = 26/52
Pred jack = 2/52
(4+26) - 2
--------- = 28/52
52

- 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

make sense?

- anonymous

yes, just going over what you are doing so I can do it on the next problem

- 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

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

Thank you for helping me and breakin it down to steps.

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