Question

Please helppp with question 1 ASAP!!!

ndative Chapter Thi °Cumulative Chapter Tw.x X aurreed . estim edamx -68 Your Account Log OutQ 尤MathLynx LIBRARY- STUDENTS INSTRUCTOAS ABOUT US CONTACT Home Colaborative Statistics Topics in Probablity Evaluation Tools Cumulative Chapter Three Evaluation Problem 1 A box has eight balls. There 1.2.3.4.5 and 6. The box is shaken. One ball i selected are two green balls and six yellow. The two green balls are numbered 1 and 2 The six yellow bails are numbered Answer the following questions, using fractions in p/q format or decimal values is selected Y that a yellow ball is selected and E that an even-numbered ball is selected accurate to the nearest 0.01 Here G wi mean that a green bal (25% Select the color/mamber pairs which comprise the sample space. bas%) P(G)- (15%) P (GIE)" L e. (15%) P(G U E) = t(15%) Are G and E mutually ex dusve? |(pick one), Problem 2

Answer 1

a) As we have 2 green balls and 6 yellow balls here, therefore the sample space here would be given as:

{ G1, G2, Y1, Y2, Y3, Y4, Y5, Y6 }

b) Probability that a green ball is selected is computed here as:

P(G) = 2 / 8 because 2 of the 8 balls are green

P(G) = 1/4

c) First we compute here:

P(E) = 4 / 8 as 4 of the 8 balls are even numbered

P(E) = 0.5

Also P(G and E) = 1 / 8 as there is one even numbered green ball

Now computing the required probability using bayes theorem, we get here:

P( G | E) = P(G and E) / P(E) = (1/8) / 0.5 = (2/8) =1 / 4

P(G | E) = 0.25

d) The required probability here is already computed in previous part as:

P(G $\cap$ E) = P(G and E) = 1 /8

P(G $\cap$ E) = 1/8

e) The required probability here is computed using addition law as:

P(G $\cup$ E) = P(G) + P(E) - P(G $\cap$ E)

P(G $\cup$ E) = (2/8) + (4/8) - (1/8)

P(G $\cup$ E) = 5/8

f) Here as we know that P(G and E) is not equal to 0, therefore G and E are not mutually exclusive