Ask question

Ask question

All categories

2 years ago

11

A bag contains 2 white marbles and 7 purple marbles. Two marbles are drawn at random. One marble is drawn and not replaced. Then

a second marble is drawn. What is the probability of selecting a purple marble and then a white marble?
A.1/7
B.2/9
C.1/14
D.7/36

1 answer:

Ghella [55]2 years ago

5 0

7/9*2/8= 7/36

your answer is D

You might be interested in

Through a point not on a line, one and only one liParallel lines are cut by a transversal such that the alternate interior angle

galina1969 [7]

<span>Parallel lines are cut by a transversal such that the alternate interior angles have measures of 3x + 17 and x + 53 degrees. The value of x is
18 </span>

5 0

2 years ago

Which statement best describes a many-to-one function?

r-ruslan [8.4K]

Answer:

B

Step-by-step explanation:

many different values of x lead to the same value of y.

hence, many-to-one.

but it does not mean that all values of x have to lead to the same value of y. it could be, but it is not demanded. so, it is not necessarily a horizontal line.

and this does not give us any information about the power of x involved. so, we cannot make any statement about linear or non-linear function.

8 0

2 years ago

What is the distance from L to M?<br> A. 10 units<br> B. 14 units<br> C. 100 units<br> D. 2 units

nignag [31]

Answer: 10 units

Explanation:

5 0

1 year ago

Quadratic shifted left by<br> 1 unit vertical stretch by a<br> factor of 1.

Aleksandr [31]

Answer:

1(x+1)^2

Step-by-step explanation:

8 0

2 years ago

How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird

NARA [144]

The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

$\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}$

Then

$Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\$

Note that since

$F \cap R=\varnothing$ , $Perm(F)\cap Perm(R)\ne \varnothing$

But since

$B \cap R \ne \varnothing$ , $Perm(B)\cap Perm(R)= \varnothing$

and

$B \cap F \ne \varnothing$ , $Perm(B)\cap Perm(F)= \varnothing$

Since

$|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\$

where $|Perm(X)|=\text{number of possible permutations of the elements of X taking all at once}$

and

$|Perm(F) \cup Perm(R)| = |Perm(F)| + |Perm(R)| - |Perm(FR)|\\= 23!+24!- 21! \text{ possibilities}$

What we are looking for is the number of permutations of the 26 letters of the alphabet that do not contain the strings fish, rat or bird, or

$|Perm(\mathcal{E})|-|Perm(B)|-|Perm(F)\cup Perm(R)|\\= 26!-23!-(23!+24!- 21!)\\= 402619359782336797900800000 \text{ possibilities}$

This link contains another solved problem on permutations:

brainly.com/question/7951365

6 0

2 years ago