Ask question

Ask question

All categories

3 years ago

13

a card is chosen at random from a deck of 52 cards. it is replaced and a second card is chosen. what is the probability of choos

ing a king and an ace?

2 answers:

serious [3.7K]3 years ago

7 0

P(2 aces) = (1/13)^2
P(2 kings) = (1/13)^2

P(king and ace) = 8C2/52C2 - 2(1/13)^2 = 0.0152

Ksju [112]3 years ago

7 0

Answer:

$\frac{1}{169}$

Step-by-step explanation:

Total cards = 52

Total ace = 4

So probability of getting ace on first draw = $\frac{4}{52}$

Now the card is replaced .

So, Total cards = 52

Total kings = 4

So probability of getting kings on second draw = $\frac{4}{52}$

So, the probability of choosing a king and an ace:

= $\frac{4}{52} \times \frac{4}{52}$

= $\frac{1}{169}$

Hence the probability of choosing a king and an ace is $\frac{1}{169}$

You might be interested in

A train is scheduled to make a journey of 300km at an average speed of 120km/hr it leaves 6 minutes late and it’s average speed

Alekssandra [29.7K]

Answer:

hsiwwbwswoksn

Step-by-step explanation:

jeiwosnn1owjsnw

6 0

3 years ago

Please simplify this fraction c:

Liula [17]

F you just want to see the really short way, just skip down to AAAAAAAAAAAA

so, here is the long explanation
exponential properties
$x^{-m}= \frac{1}{x^m}$

don't forget pemdas
2x^2=2(x^2)

so
$\frac{2x^{-4}}{3xy}$ = $\frac{2 \frac{1}{x^4} }{3xy}= \frac{2}{3x^5y}$

AAAAAAAAAAAAAAAAAAAAAAAA
so we see
the original equatio is
$\frac{2x^{-4}}{3xy}$

remember
$\frac{ab}{cd} =( \frac{a}{c})( \frac{b}{d})$

so we can seperate the constants
$\frac{ab}{cd} =( \frac{2}{3})( \frac{x^{-4}}{xy})$

we know that the placeholders cannot affet the position of the constants unless they are grouped together which they are not

terfor the answer must have 2/3 in it

the only one that hsa that is A

ANSWER IS A

3 0

3 years ago

Read 2 more answers

What fraction is 1 yard 2 feet and 6 inches

MA_775_DIABLO [31]

Answer:

Make sure you convert these numbers into one unit. Convert everything to yard. Just convert the feet and inches in yard since it's the biggest unit, and then add them together. You should get a fraction not a whole number.

3 0

3 years ago

Read 2 more answers

Help me it's algebra and dont tell me to look at the lesson :)

suter [353]

Let b be the number of blue beads and g the number of green beads that Giovanni can use for a belt.

He's supposed to use a total of between 70 and 74 beads, so

70 ≤ b + g ≤ 74

The ratio of green beads to blue beads is g/b, and this ratio has to be between 1.4 and 1.6, so

1.4 ≤ g/b ≤ 1.6

For completeness, Giovanni must use at least one of either bead color, so it sort of goes without saying that this system must also include the conditions

b ≥ 0

g ≥ 0

(These conditions "go without saying" because they are implied by the others. g/b is a positive number, so either both b and g are positive, or they're both negative. But they must both be positive, because otherwise b + g would be negative. I would argue for including them, though.)

7 0

2 years ago

Solve using long division <br> Please

madreJ [45]

1. $Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}$

$Steps:$

$\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}$

$\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}$

$\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}$

$\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}$

$\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}$

2. $Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}$

$Steps:$

$\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}$

$\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}$

$\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

4 0

3 years ago