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maybe..but then again.....
What is the probablitity of typing the word probability with no mistakes 5 times in a row?
well i have an assignment im not quite understanding.
cant read your mind...spit it out :)
her assignment is to find out what the probability is that you will randomly type the correct answer
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) =
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.
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
the probability of a red face card is 6 out of 52; (6/62) x100 for a percent
is it a red jack? or a jack or a red card?
4jack 24red cards ----- * ------------ seems right. what you think? 52 52
or it could mean you have a 26/52 chance of drawing either a red card or a jack.
would it be 6/52 or 62?
What does P(a|b) mean: a or b right?
im guessing a red jack
ya i believe so? where are you seeing that
just recalling that in programing that "|" bar means or; and after checking up on it, I was right.
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.
P(a|b) = P(a) + P(b) - P(A & B).
Pjack = 4/52 Pred = 26/52 Pred jack = 2/52 (4+26) - 2 --------- = 28/52 52
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
yes, just going over what you are doing so I can do it on the next problem
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
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"
Thank you for helping me and breakin it down to steps.