anonymous
  • anonymous
A card is chosen from a deck of cards, recorded, and then replaced. this is done 75 times and red card from 5 to 9 is chosen 21 times. a) what is the theoretical probability of a red card between 5 and 9 being chosen? b) how many times would you expect this event to happen in 75 trials? c) compare your answers to part a) and b)
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

sandra
  • sandra
so this is similar to the last one. you want to know how many possibilities there are that a given event happens (in this case red 5-9), and how many possibilities there are total. The first things you need to determine are: 1. how many red cards are there between 5 and 9? 2. how many total cards are in the deck?
anonymous
  • anonymous
well how many red cards are there in a deck .. is it 26 >?
anonymous
  • anonymous
right, so there are 5,6,7,8,9 of diamonds and hearts

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so 10 total that are red AND 5-9
anonymous
  • anonymous
there are 52 cards in the deck total
anonymous
  • anonymous
so the probability of a red card that is 5-9 = 10 (the number of red 5-9)/52 (total number of cards)
bahrom7893
  • bahrom7893
yay I get it!
anonymous
  • anonymous
how did you get 10?
anonymous
  • anonymous
well, in a deck of cards, the red cards are diamonds and hearts
anonymous
  • anonymous
so there are two sets of 5-9 that are red - hearts and diamonds
anonymous
  • anonymous
iok
anonymous
  • anonymous
so in total, there are 5,6,7,8,9 of hearts, and 5,6,7,8,9 of diamonds, so 10 total
anonymous
  • anonymous
true
anonymous
  • anonymous
so the probability = (possible red 5-9) divided by ( total possible cards)
bahrom7893
  • bahrom7893
man i hate card problems.. how the heck would i know how many cards are there...I mean gambling is illegal in schols..
bahrom7893
  • bahrom7893
*schools
anonymous
  • anonymous
lol
anonymous
  • anonymous
true - so (a) = 10/52
anonymous
  • anonymous
so how do you think you get b?
anonymous
  • anonymous
how do i rememeber all those red cards and everything?? coz itss hard :"S
anonymous
  • anonymous
well, the good news is that a deck of cards is used across a LOT of probability questions
anonymous
  • anonymous
is it lik 10/52 * 75 ?
anonymous
  • anonymous
so you just gotta know - there are 52 cards, 4 suits, 2 suits of each color, 13 cards in each suit
anonymous
  • anonymous
and yes, you are correct
bahrom7893
  • bahrom7893
YAY!
anonymous
  • anonymous
so good work =)
anonymous
  • anonymous
whoz rigjht me???
bahrom7893
  • bahrom7893
yup!
anonymous
  • anonymous
boom!
anonymous
  • anonymous
thats amazing :O
anonymous
  • anonymous
it happens :p and it will happen more the more you practice
bahrom7893
  • bahrom7893
and i still dont get it.. maybe im not tryin hard enough =/
bahrom7893
  • bahrom7893
embarassin really, bein good in calc and when someone asks me a stat question i gtg.. sorry i dunno..
anonymous
  • anonymous
haha, well if you asked me a calc question I'd give you a blank stare :p
anonymous
  • anonymous
i got aniother questin too
anonymous
  • anonymous
k, I'd say ask a different one on the left side
anonymous
  • anonymous
and I'm sure many of us will make the jump
anonymous
  • anonymous
ok make sure that uhelp me lol
anonymous
  • anonymous
ur supposed to be asking another question
bahrom7893
  • bahrom7893
shes tryin to do it on her own

Looking for something else?

Not the answer you are looking for? Search for more explanations.