Kathleen conducted an experiment in which she drew marbles from a bag one at a time and replaced the

10 answers
Question:

Kathleen conducted an experiment in which she drew marbles from a bag one at a time and replaced the marble after each draw. The results of her experiment are shown in the table.
Marble Experiment
Marble Color Number of Times
Drawn
Red 8
Blue 2
Green 6
Purple 2
Yellow 12
According to the results from Kathleen's experiment, what is the probability that she would draw a yellow marble and then draw a green marble?

Answers

150 times

Step-by-step explanation:

From the question given,

The probability of drawing a yellow triangle is 3/5 with replacement. This means that the probability of drawing a yellow card any time, will be 3/5.

For 250 draws,

The number of times a yellow triangle will appear will be = (3/5)x250 = 150

[tex]P(R,R)=\frac{3}{28}[/tex]

[tex]P(R,B)=\frac{3}{14}[/tex]

[tex]P(R,Y)=\frac{3}{56}[/tex]

[tex]P(B,R)=\frac{3}{14}[/tex]

[tex]P(B,B) =\frac{3}{14}[/tex]

[tex]P(B,Y) =\frac{1}{14}[/tex]

[tex]P(Y,R)=\frac{3}{56}[/tex]

[tex]P(Y,B) =\frac{1}{14}[/tex]

Step-by-step explanation:

Probability: The ratio of favorable outcomes to the total outcomes.

Probability usually is denoted by P(x)

Where x is favorable event.

[tex]P(x)=\frac{\textrm{favorable outcomes}}{\textrm{Total outcomes}}[/tex]

P(x,y) is the probability of choosing an x colored ball on the first drawn and a y colored ball on the second drawn.

Given that, the number of red colored ball = 3

the number of blue colored ball = 4

the number of yellow colored ball = 3

Total ball = (3+4+3)=8

The probability of drawing a red on the first drawn is

[tex]P(R)=\frac{^3C_1}{^8C_1}=\frac{3}{8}[/tex]

For second draw total number of ball = (8-1)=7

The number of red ball = (3-1)=2

The probability of drawing a red on the second drawn is

[tex]P(R)=\frac{^2C_1}{^7C_1}=\frac{2}{7}[/tex]

Therefore [tex]P(R,R)=P(R).P(R)=\frac{3}{8}. \frac{2}{7} =\frac{3}{28}[/tex]

The probability of drawing a red on the first drawn is

[tex]P(R)=\frac{^3C_1}{^8C_1}=\frac{3}{8}[/tex]

The probability of drawing a blue on the second drawn is

[tex]P(B)=\frac{^4C_1}{^7C_1}=\frac{4}{7}[/tex]

Therefore, [tex]P(R,B)=P(R).P(B)=\frac{3}{8} .\frac{4}{7} =\frac{3}{14}[/tex]

The probability of drawing a red on the first drawn is

[tex]P(R)=\frac{^3C_1}{^8C_1}=\frac{3}{8}[/tex]

The probability of drawing a yellow on the second drawn is

[tex]P(Y)=\frac{^1C_1}{^7C_1}=\frac{1}{7}[/tex]

Therefore [tex]P(R,Y)=P(R).P(Y)=\frac{3}{8}.\frac{1}{7}=\frac{3}{56}[/tex]

The probability of drawing a blue on the first drawn is

[tex]P(B)=\frac{^4C_1}{^8C_1}=\frac{4}{8}=\frac{1}{2}[/tex]

The probability of drawing a red on the second drawn is

[tex]P(R)=\frac{^3C_1}{^7C_1}=\frac{3}{7}[/tex]

Therefore [tex]P(B,R)=P(B).P(R)=\frac{1}{2} .\frac{3}{7} =\frac{3}{14}[/tex]

The probability of drawing a blue on the first drawn is

[tex]P(B)=\frac{^4C_1}{^8C_1}=\frac{4}{8}=\frac{1}{2}[/tex]

The probability of drawing a blue ball on second drawn after drawing blue ball on first drawn

Number of blue ball = (4-1)=3

[tex]P(B)= \frac{^3C_1}{^7C_1} =\frac{3}{7}[/tex]

Therefore [tex]P(B,B)=\frac{1}{2} .\frac{3}{7} =\frac{3}{14}[/tex]

The probability of drawing a blue on the first drawn is

[tex]P(B)=\frac{^4C_1}{^8C_1}=\frac{4}{8}=\frac{1}{2}[/tex]

The probability of drawing a yellow on the second drawn is

[tex]P(Y)=\frac{^1C_1}{^7C_1}=\frac{1}{7}[/tex]

Therefore [tex]P(B,Y)=\frac{1}{2} .\frac{1}{7} =\frac{1}{14}[/tex]

The probability of drawing a yellow on the first drawn is

[tex]P(Y)=\frac{^1C_1}{^8C_1} =\frac{1}{8}[/tex]

The probability of drawing a red on the second drawn is

[tex]P(R)=\frac{^3C_1}{^7C_1}=\frac{3}{7}[/tex]

Therefore [tex]P(Y,R)=\frac{1}{8}.\frac{3}{7} =\frac{3}{56}[/tex]

The probability of drawing a yellow on the first drawn is

[tex]P(Y)=\frac{^1C_1}{^8C_1} =\frac{1}{8}[/tex]

The probability of drawing a blue on the second drawn is

[tex]P(B)=\frac{^4C_1}{^7C_1}=\frac{4}{7}[/tex]

Therefore [tex]P(Y,B)=\frac{1}{8} .\frac{4}{7} =\frac{1}{14}[/tex]

Answer :

[tex]\frac{7}{15}[/tex]×[tex]\frac{2}{15}[/tex] = [tex]\frac{14}{225}[/tex]

Step-by-step explanation:

P([tex]Event_{1}[/tex]) = choosing yellow marble from paper bag = [tex]\frac{favourable outcomes}{possible outcomes}[/tex] = [tex]\frac{7}{15}[/tex]

P([tex]Event_{2}[/tex]) = choosing yellow marble from cloth bag = [tex]\frac{favourable outcomes}{possible outcomes}[/tex] = [tex]\frac{2}{15}[/tex]

∵ Resultant outcome is dependent upon both the events and both events are independent from each other, so we can apply intersection rule [ P(A∩B)=P(A)×P(B) ] here

∴ Probability ( Both marbles are yellow) = P([tex]Event_{1}[/tex]) × P([tex]Event_{2}[/tex]) = [tex]\frac{7}{15}[/tex] × [tex]\frac{2}{15}[/tex]

answer choice D would be correct 6/17

Step-by-step explanation:

NOTE THAT THERE IS NO REPLACEMENT

Step-by-step explanation:

(a)ATTACHED

(b)P(R, R) =  3/8 X 2/7 =3/28

P(R, B) =  3/8 X 4/7 = 3/14

P(R, Y) =  3/8 X 1/7 =3/56

P(B, R) =  4/8 X 3/7 =3/14

P(B, B) =  4/8 X 3/7 =3/14

P(B, Y) =  4/8 X 1/7 =1/14

P(Y, R) =  1/8 X 3/7 =3/56

P(Y, B) = 1/8 X 4/7 =1/14


[tex]There are eight balls in an urn. They are identical except for color. Three are red, four are blue,[/tex]

Omg YES !!

Explanation:

by chance is this Jade from Mortal Combat or No ? xD ( just asking )

Keep up the great work !

Sleepy~

P(B|A) =  2/9

Step-by-step explanation:

Let A be the event that the first draw  is blue marble

Let B be the event that the second draw is yellow marble.

P(B|A) = P(B∩A)/P(A)  since P(A) ≠ 0

P(B∩A) = P(B) × P(A|B)

P(B) = 6/ (3+5+4+6)

P(B) = 6/18

P(B) = 1/3

P(A|B) = 6/3

P(A|B) = 2

P(B∩A)  = 1/3 × 2

P(B∩A)  = 2/3

P(B|A) = P(B∩A)/P(A)

P(B|A) = (2/3 ) / 3

P(B|A) =  2/3 × 1/3

P(B|A) =  2/9

probably the bright blue square

The answer is "the face".


Robert Fantz began estimating the measure of time babies spent taking a gander at something as an approach to check how keen on it they were. Fantz announced that a two-month-old child spent twice as long taking a gander at a portray of the human face as at a bullseye, for example. Tests in view of look estimations have been the field's workhorse from that point onward. It is no misrepresentation to state that without looking-time measures, we would know almost no about almost any part of baby improvement.

They are so pretty. Good job

Explanation:

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