We have two urns. Urn 1 has 2 red balls and 3 yellow balls, while Urn 2 has 3 red balls and 7 yellow balls. We randomly select 75 balls form Urn 1, sampling with replacement. We independently randomly select 100 balls from Urn 2, sampling with replacement. Find the variance of the random variable defined as the number of red balls obtained from Urn 2 less the number of red balls obtained from Urn 1.

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okpalawalter8 okpalawalter8 · Answer 1 · 2021-05-13T04:15:10+02:00

Answer:

[tex]\sigma (y_1-y_2)=39[/tex]

Step-by-step explanation:

From the question we are told that:

Urn 1 :2 Red, 3 Yellow

Urn 2:3 Red, 7 Yellow

Sample 1 [tex]n_1=75\ balls[/tex]

Sample 2 [tex]n_2=100\ balls[/tex]

Generally the Probability of Red ball drawn is mathematically given by

For Urn 1

[tex]P(R)_1=\frac{2}{5}[/tex]

[tex]P(R)_1=0.4[/tex]

For Urn 2

[tex]P(R)_1=\frac{3}{10}[/tex]

[tex]P(R)_1=0.3[/tex]

Generally the equation for Variance of two independent variables [tex]\sigma (y_1-y_2)[/tex] is mathematically given by

[tex]\sigma (y_1-y_2)=\sigma y_2 +(-1)^2 \sigma(x_1)[/tex]

Where red balls drawn from both Urn is Modeled

[tex]Urn_1\ is\ Modeled\ as (n_1=75,p_1=0.4)[/tex]

[tex]Urn_2\ is\ Modeled\ as\ (n_2=100,p_2=0.3)[/tex]

[tex]\sigma (y_1-y_2)=n_2 p_2(1_p_2)+n_1*p_1(1-p_1)[/tex]

[tex]\sigma (y_1-y_2)=100*0.3(1-0.3)+75*0.4(1-0.4)[/tex]

[tex]\sigma (y_1-y_2)=39[/tex]

Therefore the variance of the random variable defined as the number of red balls is

[tex]\sigma (y_1-y_2)=39[/tex]