There are 3 bags each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles. Bag 3 has 45 red and 55 blue marbles. Now a bag is chosen at random and a marble is also picked at random. 1) What is the probability that the marble is blue? 2) What is the probability that the marble is blue when the first bag is chosen with probability 0.5 and other bags with equal probability each? Make sure to clearly define your probabilistic events and mathematically show how different probability laws and rules that you learned in class could be applied to solve the problems.

0.4 ; 0.6125

Step-by-step explanation:

Given the following :

Bag 1 : 75 red ; 25 blue

Bag 2: 60 red ; 40 blue

Bag 3: 45 red ; 55 blue

Probability = (required outcome / Total possible outcomes)

A) since the probability of choosing each bag is equal :

BAG A:

P(choosing bag A) = 1 / total number of bags = 1/3 ; P(choosing blue marble) = number of blue marbles / total number of marbles = 25/100

HENCE, choosing a blue marble from bag A : = (1/3 × 75/100) = 25/300

BAG B:

P(choosing bag B) = 1/3 ;

P(choosing blue marble) = number of blue marbles / total number of marbles = 40/100

HENCE, choosing a blue marble from bag A : = (1/3 × 40/100) = 40/300

BAG C:

P(choosing bag C) = 1/3

P(choosing blue marble) = number of blue marbles / total number of marbles = 55/100

HENCE, choosing a blue marble from bag A : = (1/3 × 55/100) = 55/300

= (25/300) × (40/300) × (55/300) = (25 + 40 + 55)/300 = 120/300 = 0.4

2) What is the probability that the marble is blue when the first bag is chosen with probability 0.5 and other bags with equal probability each?

BAG A:

P(choosing bag A) = 0.5 ; P(choosing blue marble) = number of blue marbles / total number of marbles = 25/100

HENCE, choosing a blue marble from bag A : = (0.5 × 75/100) = (0.5 * 0.75) = 0.375

BAG B:

P(choosing bag B) = (1-0.5) / 2 = 0.25 ;

P(choosing blue marble) = number of blue marbles / total number of marbles = 40/100

HENCE, choosing a blue marble from bag A : = (0.25 × 40/100) = (0.25 × 0.4) = 0.1

BAG C:

P(choosing bag C) = (1 - (0.5+0.25)) = 0.25

P(choosing blue marble) = number of blue marbles / total number of marbles = 55/100

HENCE, choosing a blue marble from bag A : = (0.25 × 55/100) = 0.25 × 0.55 = 0.1375

= 0.1375 + 0.1 + 0.375 = 0.6125

Related Questions

Find the prime factorization of
72

72 = {2,2,2,3,3}

Step-by-step explanation:

72

89

2433

22

Find the radius and height of a cylindrical soda can with a volume of 256cm^3 that minimize the surface area.B: Compare your answer in part A to a real soda can, which has a volume of 256cm^3, a radius of 2.8 cm, and a height of 10.7 cm, to conclude that real soda cans do not seem to have an optimal design. Then use the fact that real soda cans have a double thickness in their top and bottom surfaces to find the radius and height that minimizes the surface area of a real can (the surface area of the top and bottom are now twice their values in part A.

New height=?

Height: 6.88 cm.

Height: 10.92 cm.

Step-by-step explanation:

We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.

a) We can express the volume of the soda can as:

This is the constraint.

The function we want to minimize is the surface, and it can be expressed as:

To solve this, we can express h in function of r:

And replace it in the surface equation

To optimize the function, we derive and equal to zero

The radius that minimizes the surface is r=3.44 cm.

The height is then

The height that minimizes the surface is h=6.88 cm.

b) The new equation for the real surface is:

We derive and equal to zero

The radius that minimizes the real surface is r=2.73 cm.

The height is then

The height that minimizes the real surface is h=10.92 cm.

The minimal surface area for a cylindrical can of 256cm^3 is achieved with radius 3.03 cm and height 8.9 cm under uniform thickness, and radius 3.383 cm and height 7.14 cm with double thickness at top and bottom. Real cans deviate slightly from these dimensions possibly due to practicality.

Explanation:

For a cylinder with given volume, the surface area A, radius r, and height h are related by the formula A = 2πrh + 2πr^2 (if the thickness is uniform) or A = 3πrh + 2πr^2 (if the top and bottom are double thickness). By taking the derivative of A w.r.t r and setting it to zero, we can find the optimal values that minimize A.

For a volume of 256 cm^3, this gives us r = 3.03 cm and h = 8.9 cm with uniform thickness, and r = 3.383 cm and h = 7.14 cm with double thickness at the top and bottom. Comparing these optimal dimensions to a real soda can (r = 2.8 cm, h = 10.7 cm), we see that the real can has similar but not exactly optimal dimensions. This may be due to practical considerations like stability and ease of holding the can.

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What is the factors for 14 and 6

14= 1, 2, 7, 14

6= 1, 2, 3, 6

What is the greatest common factor of 5x^6+35x^4+15x^3

5 will go into everything. That is one of the factors.

x^3 is in everything as well.

The highest common factor is 5x^3

So the greatest common factor (determined by the last term) is 5x^3

And the left over polynomial is x^3 + 7x + 3

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