probability of drawing cards!? need help in a math prblm.?

Question

the question is the P(black) and P(red).
do you add them to get 1
or do you multiply to get 1/4?

and the other one is P(jack) and P(heart)
do i add to get 16/52
or multiply to get 1/52?

i'm not sure if i'm supposed to only multiply if the question is P(A) AND P(b)
or P(A) OR P(b)

Answer 1

well im not so sure about what u r asking...
when u have the word and then of course that means both. but there is a lot of information that u did not give:
Are the Jokers included?
How many times can u draw?
If the answer to the 2nd question is 1, then the 'and' must actually be an 'or' because of course u cannot pick two cards in one pick. otherwise, it will be and.
Also, im not sure what u mean with the numbers u give, but when u have probability, u must multiply. for example:
Probability of getting heads twice on a coin?
First time, the probability is 1/2.
Second time, it is also 1/2.
So, the probability of getting heads twice is (1/2)*(1/2)=1/4.
I hope that helps. if u still dont get it u r free to email me or u can try math.com.

Answer 2

probability of A AND B is P(A)•P(B).
probability of A OR B is P(A)+P(B),
provided A and B are independent.

P(black) = 1/2
P(red) = 1/2
but your question isn't clear. Are you drawing 2 cards and wanting the probability 1 is red and the other is black? In that case, any card will do for the 1st card, 52, out of 52 choices, then only 26 cards of the other color will do out of 51 choices: (52•26)/(52•51) = 26/51 = 0.5098

you're too loose with "and" in this context. prob(jack) = 4/52 = 1/13, and prob(heart) = 13/52 = 1/4. so prob(jack and heart) = prob(jack OF hearts) = (1/4)(1/13) = 1/52, since there's only 1 such card.

prob(jack OR heart) = 1/13 + 1/4 - 1/52 = 4/52 + 13/52 - 1/52 = 16/52 = 4/13. you subtract the 1/52 so as not to count the jack of hearts twice, since it's 1 of the 4 jacks and also 1 of the 13 hearts.

Answer 3

if events are independent, then the probability of both happening is the product. So multiply for AND

So P(heart) = 1/4, P(jack) = 1/13
therefore P(jack AND heart) = 1/4 * 1/13 = 1/52 which makes sense since there is one jack of hearts in the 52-card deck.
Now if you were asked what the chances are of getting a heart or a jack, the answer is
P(heart) + P(jack) - P(jack AND heart) since the jack of hearts would be counted in both of the first two probabilities.
P(heart) = 1/4, P(jack) = 1/13, P(jack AND heart) = 1/52
1/4 + 1/13 - 1/52 = 16/52 = 4/13

Answer 4

Please clarify: Are you drawing 2 cards or 1?

If you are drawing 1 card, then:
P(Jack of hearts) = P(Jack) * P(heart) = 1/13 * 1/4 = 1/52.
P(Jack OR heart) = P(Jack) + P(heart) = 1/13 + 1/4 = 16/52.
P(black AND red) = 0. You can't draw a card that is both red and black.
P(black OR red) = 1. A card is either red or black.

If you are drawing 2 cards, then the probabilities are different and more complicated.

Answer 5

it seems to me it is seen like this: leaf with the help of the deck, come across 2 spades and located them aside. Shuffle the deck and draw an extra constructive card at random. What are the possibilities it is a spade ? eleven spades left / 50 enjoying cards left = eleven/50. so ordinary as that. it is not any longer any different than "what are the possibilities the 0.33 card is a spade given the 1st 2 are spades ?". if so it is clearer that the region above is equivalent (intentionally relatively than randomly removing 2 spades up the front), on account which you're _given_ that 2 of the enjoying cards are spades. .