Consider 5 boxes, each of which contains 4 balls; in particular, box 1 contains 4 white balls, box 2 contains 3 white balls and 1 red ball, box 3 contains 2 white balls and 2 red balls, box 4 contains 1 white ball and 3 red balls and box 5 contains 4 red balls. I choose a box at random and I extract 2 balls with replacement. Show that the probability that the 2 balls are red is equal to 3/8.

The probability of getting 2 red balls is 0.68.

Step-by-step explanation:

College Mathematics 5+3 pts

Urn 1 contains 5 red balls and 3 black balls. Urn 2 contains 3 red balls and 1 black ball. Urn 3 contains 4 red balls and 2 black balls. If an urn is selected at random and a ball is drawn, find the probability it will be red.

Report 04.10.2016

Answers

caylus

CaylusAmbitious

Hello,

Choice urn_1: 1/3

choice R: 5/8 ==>1/3*5/8=5/24

Choice urn_2: 1/3

Choice R: 3/4 ==> 1/3*3/4=1/4

Choice urn_3: 1/3

Choice R:4/6=2/3 ==> 1/3*2/3=2/9

Total: 5/24+1/4+2/9=15/72+18/72+16/72=49/72

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PinquancaroAmbitious

The probability of getting red balls is 0.68.

Step-by-step explanation:

Given : Urn 1 contains 5 red balls and 3 black balls. Urn 2 contains 3 red balls and 1 black ball. Urn 3 contains 4 red balls and 2 black balls. If an urn is selected at random and a ball is drawn.

To find : The probability it will be red ?

Solution :

Total urn = 3

If an urn is selected at random then the probability is P(U)=\frac{1}{3}

In urn 1 - 5 red balls + 3 black balls

Probability of getting red ball from urn 1 - P(R_1)=\frac{5}{8}

In urn 2 - 3 red balls + 1 black balls

Probability of getting red ball from urn 2 - P(R_2)=\frac{3}{4}

In urn 3 - 4 red balls + 2 black balls

Probability of getting red ball from urn 3 - P(R_3)=\frac{4}{6}

Choosing red ball from urn is P(R)=P(R_1)+P(R_2)+P(R_3)

P(R)=\frac{5}{8}+\frac{3}{4}+\frac{4}{6}

P(R)=\frac{15+18+16}{24}

P(R)=\frac{49}{24}

The probability of getting red balls is

P=P(U)\times P(R)

P=\frac{1}{3}\times \frac{49}{24}

P=\frac{49}{72}

P=0.68

We are doing this with replacement.

The probability of choosing any box = 1/5

Box 1: 4 white + 0 red. P(getting red) = 0. Therefore P(choosing box 1 and 2 red balls) = 1/5 * 0 = 0

Box 2: 3 white + 1 red. P(getting red) = 1/4. Therefore P(choosing box 2 and 2 red balls) = 1/5 * 1/4 * 1/4 = 1/80

Box 3: 2 white + 2 red. P(getting red) = 2/4. Therefore P(choosing box 3 and 2 red balls) = 1/5 * 2/4 * 2/4 = 4/80

Box 4: 1 white + 3 red. P(getting red) = 3/4. Therefore P(choosing box 4 and 2 red balls) = 1/5 * 3/4 * 3/4 = 9/80

Box 5: 0 white + 4 red. P(getting red) = 4/4. Therefore P(choosing box 5 and 2 red balls) = 1/5 * 4/4 * 4/4 = 16/80

Therefore the required probability = 1/80 + 4/80 + 9/80 + 16/80 = 30/80 = 3/8

Step-by-step explanation:

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841

ur mom

step-by-step explanation: