1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
enyata [817]
3 years ago
6

The following exercise refers to choosing two cards from a thoroughly shuffled deck. Assume that the deck is shuffled after a ca

rd is returned to the deck. If you do not put the first card back in the deck before you draw the next, what is the probability that the first card is a 4 and the second card is a ace? (Enter your probability as a fraction.)
Mathematics
2 answers:
dem82 [27]3 years ago
8 0

Answer:

\frac{4}{663}

Step-by-step explanation:

Given that from a well shuffled set of playing cards (52 in number) a card is drawn and without replacing it, next card is drawn.

A - the first card is 4

B - second card is ace

We have to find probability for

A\bigcap B

P(A) = no of 4s in the deck/total cards = \frac{4}{52} =\frac{1}{13}

After this first drawn if 4 is drawn, we have remaining 51 cards with 4 aces in it

P(B) = no of Aces in 51 cards/51 = \frac{4}{51}

Hence

P(A\bigcap B) = \frac{1}{13} *\frac{4}{51} \\=\frac{4}{663}

(Here we see that A and B are independent once we adjust the number of cards. Also for both we multiply the probabilities)

siniylev [52]3 years ago
8 0

Answer: The probability is P =  4/663

Step-by-step explanation:

We have two events:

A: Drawing a 4

B: Drawing an ace.

If each card has the same probability to being selected, then:

For the first event, we can calculate the probability as the number of 4's in the deck divided the total number of cards in the deck

We have 4 4's, and 52 total cards, so the probability is:

Pa = 4/52 = 1/13

Now, if we do not return the first card to the deck now we have 51 cards in the deck.

Now the probability of drawing an ace is equal to the number of aces in the deck divided the total number of cards in the deck, this is:

Pb = 4/51.

The joint probability of both events happening is equal to the product of their probabilities, this is:

P = Pa*Pb = (1/13)*(4/51) = 4/(13*51) = 4/663

You might be interested in
Evaluate the limit
wel

We are given with a limit and we need to find it's value so let's start !!!!

{\quad \qquad \blacktriangleright \blacktriangleright \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

But , before starting , let's recall an identity which is the <em>main key</em> to answer this question

  • {\boxed{\bf{a^{2}-b^{2}=(a+b)(a-b)}}}

Consider The limit ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

Now as directly putting the limit will lead to <em>indeterminate form 0/0.</em> So , <em>Rationalizing</em> the <em>numerator</em> i.e multiplying both numerator and denominator by the <em>conjugate of numerator </em>

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}\times \dfrac{\sqrt{x}+\sqrt{3\sqrt{x}-2}}{\sqrt{x}+\sqrt{3\sqrt{x}-2}}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-\sqrt{3\sqrt{x}-2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{(x^{2}-4^{2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Using the above algebraic identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x})^{2}-(\sqrt{3\sqrt{x}-2})^{2}}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-(3\sqrt{x}-2)}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}+2}{\{(\sqrt{x})^{2}-2^{2}\}(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , here we <em>need</em> to <em>eliminate (√x-2)</em> from the denominator somehow , or the limit will again be <em>indeterminate </em>,so if you think <em>carefully</em> as <em>I thought</em> after <em>seeing the question</em> i.e what if we <em>add 4 and subtract 4</em> in <em>numerator</em> ? So let's try !

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2+4-4}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(x-4)+2+4-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , using the same above identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+6-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+3(2-\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take minus sign common in <em>numerator</em> from 2nd term , so that we can <em>take (√x-2) common</em> from both terms

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)-3(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take<em> (√x-2) common</em> in numerator ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)\{(\sqrt{x}+2)-3\}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Cancelling the <em>radical</em> that makes our <em>limit again and again</em> <em>indeterminate</em> ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\cancel{(\sqrt{x}-2)}\{(\sqrt{x}+2)-3\}}{\cancel{(\sqrt{x}-2)}(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}+2-3)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-1)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , <em>putting the limit ;</em>

{:\implies \quad \sf \dfrac{\sqrt{4}-1}{(\sqrt{4}+2)(4+4)(\sqrt{4}+\sqrt{3\sqrt{4}-2})}}

{:\implies \quad \sf \dfrac{2-1}{(2+2)(4+4)(2+\sqrt{3\times 2-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{6-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{4})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+2)}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(4)}}

{:\implies \quad \sf \dfrac{1}{128}}

{:\implies \quad \bf \therefore \underline{\underline{\displaystyle \bf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}=\dfrac{1}{128}}}}

3 0
2 years ago
Read 2 more answers
PLEASE HELP!
Kay [80]

Answer:

77 inches

Step-by-step explanation:

Given

1 inch = 3 miles in map

We are also given the actual distance between City A and City B = 231 miles.

In order to get the distance in inches = Distance between City A and City B in miles/ 3

We are dividing the distance by 3 to get the distance in inches.

So,

Distance on map = 231/3

= 77 inches

4 0
4 years ago
Read 2 more answers
Why are all the questons just for 5 points no one choses higher
ValentinkaMS [17]

Answer:

I don't know I want even let me choose 5

3 0
3 years ago
What is the equation of the line which includes points (1, 1) and (1, 5)?
Allisa [31]
Equation of the line that includes points (1, 1) and (1, 5) is a vertical line given by x = 1.
6 0
3 years ago
Find the equation of the line which passes through the point
sammy [17]

Answer:

y= \frac{1}{5}x- \frac{56}{5}

Step-by-step explanation:

We want to find the equation of the line that goes through: !6,-10) and is perpendicular to the line

5x + 3y = 2y - 3

Let's simplify the equation of the line to get:

3y - 2y = - 5x + 3

y = - 5x + 3

The slope of this line is -5.

The perpendicular to this line has a slope of

\frac{1}{5}

The equation going through (6,-10) is

y=m(x-x_1)+y_1

We substitute to get;

y= \frac{1}{5} (x-6) - 10

Expand;

y= \frac{1}{5}x- \frac{6}{5}  - 10

y= \frac{1}{5}x- \frac{56}{5}

7 0
3 years ago
Other questions:
  • 2. Describe the process used to solve an equation with variables on both sides.
    6·1 answer
  • Which point of intersection is the solution to the system that of equations y= 2/5x-1/2 and y=-1/3x+2/3
    9·1 answer
  • A roast beef weighs 5.65 kilograms. How grams does it weigh?
    14·1 answer
  • Factor 4x^2+28+49<br> A.(x+7)(4x+7)<br> B.4(x+7)(x+7)<br> C.(2x+7(2x+7)<br> D.2(x+7)(x+7)
    8·1 answer
  • carl used 2 1/2 cup of peanuts for every 1/2 cups of raisins when making a mix. enter the number of cups of peanuts carl uses fo
    9·2 answers
  • At a certain airport, 65% of the flights arrive on time. A sample of 10 flights is studied. Find the probability that all 10 of
    13·1 answer
  • Whats the area of each figure ?
    13·1 answer
  • Help plzz! True or False? <br> -15 is a solution for -8x = 125
    6·2 answers
  • Emily has $19 more dollars than Juan. Juan has $65. How much money does Emily have? Type your answer in the box below​
    13·2 answers
  • Question 2 of 10
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!