Question

There are eight balls in an urn. They are identical except for color. Three are red, four are blue, and one is yellow. You are to draw a ball from the urn, note its color, and set it aside. Then you are to draw another ball from the urn and note its color. Let P(x, y) be the probability of choosing an x-colored ball on the first draw and a y-colored ball on the second draw. Compute the probability for each outcome of the experiment. (Enter your answers as fractions.)P(R, R) = _____P(R, B) = _____P(R, Y) = _____P(B, R) = _____P(B, B) = _____P(B, Y) = _____P(Y, R) = _____P(Y, B) = _____

Answer 1

Answer:

$P(R,R)=\frac{3}{28}$

$P(R,B)=\frac{3}{14}$

$P(R,Y)=\frac{3}{56}$

$P(B,R)=\frac{3}{14}$

$P(B,B) =\frac{3}{14}$

$P(B,Y) =\frac{1}{14}$

$P(Y,R)=\frac{3}{56}$

$P(Y,B) =\frac{1}{14}$

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.

$P(x)=\frac{\textrm{favorable outcomes}}{\textrm{Total outcomes}}$

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

$P(R)=\frac{^3C_1}{^8C_1}=\frac{3}{8}$

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

$P(R)=\frac{^2C_1}{^7C_1}=\frac{2}{7}$

Therefore $P(R,R)=P(R).P(R)=\frac{3}{8}. \frac{2}{7} =\frac{3}{28}$

The probability of drawing a red on the first drawn is

$P(R)=\frac{^3C_1}{^8C_1}=\frac{3}{8}$

The probability of drawing a blue on the second drawn is

$P(B)=\frac{^4C_1}{^7C_1}=\frac{4}{7}$

Therefore, $P(R,B)=P(R).P(B)=\frac{3}{8} .\frac{4}{7} =\frac{3}{14}$

The probability of drawing a red on the first drawn is

$P(R)=\frac{^3C_1}{^8C_1}=\frac{3}{8}$

The probability of drawing a yellow on the second drawn is

$P(Y)=\frac{^1C_1}{^7C_1}=\frac{1}{7}$

Therefore $P(R,Y)=P(R).P(Y)=\frac{3}{8}.\frac{1}{7}=\frac{3}{56}$

The probability of drawing a blue on the first drawn is

$P(B)=\frac{^4C_1}{^8C_1}=\frac{4}{8}=\frac{1}{2}$

The probability of drawing a red on the second drawn is

$P(R)=\frac{^3C_1}{^7C_1}=\frac{3}{7}$

Therefore $P(B,R)=P(B).P(R)=\frac{1}{2} .\frac{3}{7} =\frac{3}{14}$

The probability of drawing a blue on the first drawn is

$P(B)=\frac{^4C_1}{^8C_1}=\frac{4}{8}=\frac{1}{2}$

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

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

$P(B)= \frac{^3C_1}{^7C_1} =\frac{3}{7}$

Therefore $P(B,B)=\frac{1}{2} .\frac{3}{7} =\frac{3}{14}$

The probability of drawing a blue on the first drawn is

$P(B)=\frac{^4C_1}{^8C_1}=\frac{4}{8}=\frac{1}{2}$

The probability of drawing a yellow on the second drawn is

$P(Y)=\frac{^1C_1}{^7C_1}=\frac{1}{7}$

Therefore $P(B,Y)=\frac{1}{2} .\frac{1}{7} =\frac{1}{14}$

The probability of drawing a yellow on the first drawn is

$P(Y)=\frac{^1C_1}{^8C_1} =\frac{1}{8}$

The probability of drawing a red on the second drawn is

$P(R)=\frac{^3C_1}{^7C_1}=\frac{3}{7}$

Therefore $P(Y,R)=\frac{1}{8}.\frac{3}{7} =\frac{3}{56}$

The probability of drawing a yellow on the first drawn is

$P(Y)=\frac{^1C_1}{^8C_1} =\frac{1}{8}$

The probability of drawing a blue on the second drawn is

$P(B)=\frac{^4C_1}{^7C_1}=\frac{4}{7}$

Therefore $P(Y,B)=\frac{1}{8} .\frac{4}{7} =\frac{1}{14}$