probibility?

Question

You have a standard deck of 52 playing cards with 4 suits. If two cards are drawn, what is the chance of selecting two queens?

Answer 1

Probability of drawing the first queen are 4/52 or 1/13.

Probability of drawing a second queen is then 3/51 or 1/17

So the probabilty of both events happening is the probability of both events multiplied, ie:

1/13 x 1/17 = 1 / 221

Answer 2

the first card you draw has a 4/52 chance of being a queen.

Assuming you drew a queen with the first card, the chances of the second card being a queen are now 3/51. Multiply these together for your answer:

4/52 * 3/51 = 12/2652 = 0.45% chance or less than 1/200.
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Couple other interesting scenarios:

What are the chances of drawing two cards that are the same?
This time it doesn't matter what the first card was, all you have to do is match the second card to the first.
Therefore 3/51 = 5.88%

What are the chances of drawing a queen with either of your two cards. The way to do this is to determine the odds of -not- drawing a queen with either card and then subtracting from 1.
Odds of not drawing a Q with first card: 48/52
Odds of not drawing a Q with second card: 48/51
Odds of not drawing a Q with either card: (48*48)/(51*52)

Odds of drawing a queen with either of the two cards:
1 - (48*48)/(51*52) = 1 - 86.88% = 13.12%

Answer 3

the probability for drawing first queen is 4 out of 52
= 4/52 that's = 1/13
and the probability of drawing second queen is 3/51
= 3/51 that's = 1/17

the probability of drawing 2 queens would be 1/13*1/17=1/221

Answer 4

1 in 20

Answer 5

52 x 52

Answer 6

P(probability of two queen)=(4/52)x(3/51) =12/2652

Answer 7

(4/52)(3/51) = (1/13)(1/17) = (1/221)

Answer 8

there are 4 queens , chance of first being queen

P(4,52) + secod beig queen P(3,52)
since oe queen is already selected(draw) we have oly 3 queens left .