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can you give an example of what type of problem you're having difficulty with?
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?
pleaaw help meeee
so what are the possible outcomes for (a) after the two coins are tossed?
2 over 4 ??
what is the prob that one coin will land on heads?
i dont know thats y im asking
1/2 for one coin right? has to be either heads or tails.
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)
oh okay !! :)
and C = 1/4 right?
think about your possible outcomes for one coin being heads? the possible outcomes of the coin toss are: HH, TT, HT, and TH.
prob for Only* one being heads is 1:2, if you count HH as one being heads, yes 3:4.
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
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 !
starting with a: how many red cards in a deck of cards?
half of the deck is red (diamonds and hearts) = 26/52, so prob of selecting a red card is 26/52 = 1/2
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
oo man :(
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
yay i got that right :)
good, starting to make sense?
c) im not sure is it?? 24/52?
c) there are 5 even numbers for each of the 4 suits (2, 4, 6,8, 10), 20 in total, so 20/52.
yes, do you understand?
yes i doo :D
ok, great ; )
i want you to help me for 1 more question can i help mme?
part d) or what is the question?
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 ?
ok, so you can find the area of each geometric shape (the square and triangle) right?
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
b) prob that connie did not hit triangle = (area of square-area of triangle)/(area of square) = (400-40)/(400) = 360/400 = 9/10
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).
Still make sense? Probability at this level is just the likeness of an event taking place.
Goodluck ; )