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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Report

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:

d

step-by-step explanation:

given

y + 3 = - 4(x - 2) ← distribute the parenthesis

y + 3 = - 4x + 8 ( subtract 3 from both sides )

y = - 4x + 5 → d

just needed some points but try socratic