anonymous
  • anonymous
i need help with probabilities ..can anyone please help me?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
can you give an example of what type of problem you're having difficulty with?
anonymous
  • anonymous
Two coins are tossed at the same time. what is the theoretical probability that: a) both are heads ? b) neither are heads? c) one is heads?
anonymous
  • anonymous
pleaaw help meeee

Looking for something else?

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

More answers

anonymous
  • anonymous
so what are the possible outcomes for (a) after the two coins are tossed?
anonymous
  • anonymous
2 over 4 ??
anonymous
  • anonymous
hello?
anonymous
  • anonymous
what is the prob that one coin will land on heads?
anonymous
  • anonymous
i dont know thats y im asking
anonymous
  • anonymous
1/2 for one coin right? has to be either heads or tails.
anonymous
  • anonymous
yes
anonymous
  • anonymous
so, the prob that both coins are heads is the prob that one coin landing on heads multiplied by the prob that the second coin lands on heads: (.5) * (.5) = .25 or (1/4)
anonymous
  • anonymous
oh okay !! :)
anonymous
  • anonymous
and C = 1/4 right?
anonymous
  • anonymous
think about your possible outcomes for one coin being heads? the possible outcomes of the coin toss are: HH, TT, HT, and TH.
anonymous
  • anonymous
3/4
anonymous
  • anonymous
prob for Only* one being heads is 1:2, if you count HH as one being heads, yes 3:4.
anonymous
  • anonymous
I'm not sure how far you've gotten into prob, but it's called a term call union: the prob of P(1h) U P(2h) = 3:4
anonymous
  • anonymous
A card is selescted from a deck of cards. fine the theoretical probability of each outcome. a) a red face card b) a blackface card that is a male ( king or jack) c) a numbered card that is an even number ( Hint : an ace is an odd card) d) ad odd spade e)a red ace !
anonymous
  • anonymous
starting with a: how many red cards in a deck of cards?
anonymous
  • anonymous
yes 26
anonymous
  • anonymous
half of the deck is red (diamonds and hearts) = 26/52, so prob of selecting a red card is 26/52 = 1/2
anonymous
  • anonymous
I'm not sure what they mean by face card, if by that they mean kings, queens, jacks, that will change the answer because there are two red kings, two red queens, two red jacks, so 6/52 = 3/26
anonymous
  • anonymous
oo man :(
anonymous
  • anonymous
likewise for b) a black facecard that is a male (king or jack): so there are two black kings and two black jacks, so 4/52 = 2/26 = 1/13
anonymous
  • anonymous
yay i got that right :)
anonymous
  • anonymous
good, starting to make sense?
anonymous
  • anonymous
yes :)
anonymous
  • anonymous
c) im not sure is it?? 24/52?
anonymous
  • anonymous
c) there are 5 even numbers for each of the 4 suits (2, 4, 6,8, 10), 20 in total, so 20/52.
anonymous
  • anonymous
okay
anonymous
  • anonymous
hello ?
anonymous
  • anonymous
yes, do you understand?
anonymous
  • anonymous
yes i doo :D
anonymous
  • anonymous
ok, great ; )
anonymous
  • anonymous
i want you to help me for 1 more question can i help mme?
anonymous
  • anonymous
ok
anonymous
  • anonymous
part d) or what is the question?
anonymous
  • anonymous
Connie made a square dart board with a side length of 20cm. inside the square is a triangle with base 10 cm and height 8cm. she throws a dart and hits sosmewhere within the square. : a) what is the probability that connie hit the triangle? b) what is the probability that connie did not hit the triangle ? c) should your answers to parts a ) and b) add to 1 ? explain ?
anonymous
  • anonymous
ok, so you can find the area of each geometric shape (the square and triangle) right?
anonymous
  • anonymous
area of square = s * s = 20 * 20 = 400cm^2 area of triangle = (1/2)b * h = (1/2)10 * 8 = 40cm^2 so for a) prob that connie hit the triangle = (area of triangle)/(area of square) = 40cm^2/400cm^2 = 1/10
anonymous
  • anonymous
b) prob that connie did not hit triangle = (area of square-area of triangle)/(area of square) = (400-40)/(400) = 360/400 = 9/10
anonymous
  • anonymous
c) yes, for this particular problem given the information, answers from part a & part b should add to equal 1, because when the dart lands in the square, it MOST be either in the triangle whose area represents (1/10) of the square or outside of the triangle but still inside the square, where the area reapresents (9/10).
anonymous
  • anonymous
Still make sense? Probability at this level is just the likeness of an event taking place.
anonymous
  • anonymous
Goodluck ; )

Looking for something else?

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