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eReport IMaeLy xCurmiati e Oapter Thn acumulative Chapter Twi x x auread . estim Your Account Log OutQSte Search MathLynx LIBRARY STUDENTS ADOUT US Home Collaborative Statistics Topcs in Probability Evaluation Tools Cumulative Chapter Three Evaluation Problem 1 A box has eight balls. There are six green balls and twe yelow. The six green balls are numbered 1 2,3.4.5.and 6 are numbered 1 and 2 The box is shaken One ballis selected. The two yellow bals Here G Answer the following questions, using fractions in p/q format or decimall values accurate to the nearest 0.01 s selected. Y that a yellow ball is selected and E that an even-numbered ball is selecterd ill mean that a green ball (2596) Select the color/number pairs which comprise the sample space. 1(15%) P ( E)- E mutually exclusive? [(pick one) t(15%) Are G and Problem2

Explanation / Answer

Problem 1

Part (a)

The sample space: {G1, G2, G3, G4, G5, G6, Y1, Y2} ANSWER [ since there are 6 green balls numbered 1 to 6 and 2 yellow balls numbered 1 and 2]

Part (b)

Out of 8 balls, 6 are green. So, P(G) = 6/8 = 0.75 ANSWER

Part (c)

Out of 8 balls, 4 are even numbered, i.e., G2, G4, G6, Y2; and of these 4, three are green. So, P(G/E) = 6/8 = 0.75 ANSWER

Part (d)

Out of 8 balls, 3 are green and even numbered, i.e., G2, G4, G6.

So, P(G E) = 3/8 = 0.375 ANSWER

Part (e)

Out of 8 balls, 7 are green or even numbered, i.e., G1, G2, G3, G4, G5, G6, Y2.

So, P(G E) = 7/8 = 0.875 ANSWER

Part (f)

Out of 8 balls, since 6 are green, P(G) = 6/8 and 4 are even numbered, i.e., G2, G4, G6, Y2 implying P(E) = 4/8. So, P(G) x P(E) = 24/64 = 3/8 = 0.375.

Now, P(G E) = 3/8 = 0.375 [see Part (d)]    

Thus, P(G E) = P(G) x P(E) => G and E are independent ANSWER

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