Are drawn together from a pack of 52 cards the probability that one is a spade and one is a heart is?

because there is no way to draw 2 cards independently unless you draw one, replace it and reshuffle the deck and draw the other.

Why do you think the first answer requires independence? If we did it with independence the probability the first is a heart then a a spade would be $\frac 14\cdot \frac 14$ and the probability of a spade then a heart would be $\frac 14\cdot \frac 14$ and the probability would be $\frac 14 \frac 14 + \frac 14 \frac 14 = \frac 18$.

$P(S \cap H) = P(S)P(H | S) = \frac{1}{4} \times \frac{13}{51} = \frac{13}{204}.$

That's not quite right. What does "$S$" mean? Does it mean a specific card is a spade? Or does it mean at least one of two cards is a spade? Or something else. If $P(S) = \frac 14$ this seems to imply you mean a specific card is a spade. But then $P(S\cap H)$ would mean the probability that a specific card is both a heart and a spade and $P(H|S)$ is the probability of a card being a spade given we know it is a heart. (So $0 = P(S\cap H) = P(S)P(H|S) = \frac 14\cdot 0 = 0$.)

Or maybe $S$ means a specific card is a spade, and $H$ means the other card is a heart. Then your calculation is correct..... But you figured the probability that a specific card is a spade and the other specific card is a heart.... And that was not the question. The question was that either card is a spade and the other card is a heart.

If $S$ at least one card of two is a spade and $H$ is at least one card of two is a heart then to calculate conditional probability would go like this:

$P(H\cap S) = P(S)P(H|S)$ is $\frac {13*39 + 39*13+13*13}{52*51}\cdot \frac{ 13*13 + 13*13}{13*39 + 39*13 + 13*13}=$

$\frac {13\cdot 26}{52*51}= \frac {13}{102}$

But that's a ridiculously hard way to do it.

Better to either figure there are $2\times 13 \times 13$ (heart, spade) and (spade, heart) pairs where order matters out of $52\times 51$ combos; or there are $13\times 13$ (heart,spade) pairs were order doesn't matter out of ${52\choose 2}$ combos.

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tl;dr

you figured out the probability of a specific card being a spade and the other not being. As order does not matter the probability is one half of that.

Exercise :: Probability - General Questions

  • Probability - Important Formulas
  • Probability - General Questions

11. 

A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:

A.
B.
C.
D.

Answer: Option C

Explanation:

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

Are drawn together from a pack of 52 cards the probability that one is a spade and one is a heart is?
P(E) =
n(E) = 2 = 1 .
n(S) 52 26

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12. 

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

A.
B.
C.
D.

Answer: Option C

Explanation:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15
= 15C3
= (15 x 14 x 13)
(3 x 2 x 1)
= 455.

Let E = event of getting all the 3 red balls.

Are drawn together from a pack of 52 cards the probability that one is a spade and one is a heart is?
n(E) = 5C3 = 5C2 =
(5 x 4) = 10.
(2 x 1)

Are drawn together from a pack of 52 cards the probability that one is a spade and one is a heart is?
P(E) =
n(E) = 10 = 2 .
n(S) 455 91

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Page 2

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),      (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),

     (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Are drawn together from a pack of 52 cards the probability that one is a spade and one is a heart is?
n(E) = 27.

Are drawn together from a pack of 52 cards the probability that one is a spade and one is a heart is?
P(E) =
n(E) = 27 = 3 .
n(S) 36 4

Q:

Find the range of the data 2, 1, 2, 3, 5, 4, 7, 3, 5, 2, 4.

Answer & Explanation Answer: D) 6

Explanation:


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