Ask question

Ask question

All categories

2 years ago

11

A bag contains 3 red marbles and 8 blue marbles. A marble is drawn from the bag, the color recorded, then the marble is put back

into the bag. This process is repeated ten times, and we are interested in the number of blue marbles drawn. What would be the probability of success in this experiment

1 answer:

Oksana_A [137]2 years ago

3 0

Answer:

0.0414

Step-by-step explanation:

Total number of marbles= 8+3= 11

8 blue and 3 red

Marbles are picked with replacement until 10 marbles are selected.

Probability of number of blue marbles choosen in 10 pick with replacement out of 11 marbles= ((8C1/11C1))^10

=(8/11)^10

=0.72723^10

= 0.0414

You might be interested in

Is this correct? Pls help within 10 minutes will mark brainliest

Vedmedyk [2.9K]

Answer:

the first answer is correct

the second answer is correct if they messed up on the question and meant to ask for the MAD of set B

I think the 3rd answer is incorrect (I got 1.09), so you might want to double-check your 3rd answer since it is asking for "times bigger", not how much bigger, which means how many times larger is the MAD of set A than that of set B and to find how many times bigger a number is than another number, you have to divide the first number by the second number.

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Andrew was making a science project for school. He worked for 6.6 hours the first week, 2.75 hours the second week, and 4.4 hour

fiasKO [112]

Answer:

Step-by-step explanation:

its wierd but 2.22

8 0

2 years ago

Let X be the number of the cars being repaired at a repair shop. We have the following information:

worty [1.4K]

Answer:

(a) Sample Space

$S = \{0,1,2,3\}$

(b) PMF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

(c) CDF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

$S = \{0,1,2,3\}$

Solving (b): The PMF

From the question, we have:

$P(2) = P(1)$

$P(0) = P(3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$ can be represented as:

$P(1) + P(2) = 0.5[P(0) + P(3)]$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(1) + P(1) = 0.5[P(0) + P(0)]$

$2P(1) = 0.5[2P(0)]$

$2P(1) = P(0)$

$P(0)= 2P(1)$

Also note that:

$P(0) + P(1) + P(2) + P(3) = 1$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(0) + P(1) + P(1) + P(0) = 1$

$2P(1) + 2P(0) = 1$

Substitute $P(0)= 2P(1)$

$2P(1) + 2*2P(1) = 1$

$2P(1) + 4P(1) = 1$

$6P(1) = 1$

Solve for P(1)

$P(1) = \frac{1}{6}$

To calculate others, we have:

$P(2) = P(1)$

$P(2) = P(1) = \frac{1}{6}$

$P(0)= 2P(1)$

$P(0) =2 * \frac{1}{6}$ $P(0) =\frac{1}{3}$

$P(3) = P(0) =\frac{1}{3}$

Hence, the PMF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

<em>See attachment (1) for histogram</em>

Solving (c): The CDF ; F(x)

This is calculated as:

$F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)$

For x = 0;

We have:

$P(X \le 0) = P(0)$

$P(X \le 0) = 1/3$

For x = 1

$P(X \le 1) = P(0) + P(1)$

$P(X \le 1) = 1/3 + 1/6$

$P(X \le 1) = 1/2$

For x = 2

$P(X \le 2) = P(0) + P(1) + P(2)$

$P(X \le 2) = 1/3 + 1/6 + 1/6$

$P(X \le 2) = 2/3$

For x = 3

$P(X \le 3) = P(0) + P(1) + P(2) + P(3)$

$P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3$

$P(X \le 3) = 1$

Hence, the CDF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

<em>See attachment (2) for histogram</em>

6 0

2 years ago

Is this a function? Explain

Andru [333]

Answer:

No

Step-by-step explanation:

This is not a function because there is no y and x intercept in the following graph as y=0

4 0

2 years ago

Read 2 more answers

The function f(x) = 1(3)* models the number of virus cells present in a

swat32

Answer:

$\lim_{n \to \infty} a_n \int\limits^a_b {x} \, dx$

Step-by-step explanation:

4 0

2 years ago