4. You have a jar that contains black marbles and gold marbles. There are a tota

Question

4. You have a jar that contains black marbles and gold marbles. There are a total of 10 marbles in the jar. A trial consists of taking out a marble at random and then replacing it. You conduct a total of m trials. The expected number of black marbles you observe during the m trials is 2.8 and the variance of the number of black marbles you observe during the m trials is 0.84.

(a) Determine m and the composition of marbles in this jar. You then put the marbles back to the jar. Your friend comes along. She conducts the same trials, and records the number of trials she observes the first gold marble.

(b) What is the probability that she observes the first gold marble on the fifth trial given that the first gold marble appears on an odd-numbered trial?

(c) What is the variance of the number of trials on which you observe the first gold marble?

Explanation / Answer

Total number of marbles = 10

Here number of trials = m

Here expected number of black marbles in m trials = 2.8

and variance = 0.84

That means if there are k black marbles out of 10 so Pr(Black Marbles) = k/10

and

m * k/10 = 2.8

mk = 28

Variance of the number of black marbles = (k/10) * (10-k)/10 * m = 0.84

mk(10-k) = 84

so, 10-k = 3

k = 7

and m = 4

so number of trials = 4 and number of black marbles = 7

(b) Here we have to find that

Pr(First gold marbe on 5th trial l first gold marble on an odd - numbered trial)

Pr(First gold marble on an odd numbered trial) = Pr(Gold marble on first attempt) + Pr(Gold marble on third attempt) + .....

= 0.3 + (1-0.3)2 * 0.3 + (1-0.3)4 * 0.3 + ....

= 0.3 [ 1 + 0.72 + 0.74 + ....]

= 0.3/ ( 1- 0.72) = 0.3/0.51 = 30/51 = 10/17

Pr(First gold marble on fifth trial) = 0.74* .3  

Pr(First gold marbe on 5th trial l first gold marble on an odd - numbered trial) = 0.74 * 0.3/ (10/17) = 0.1225

(c) Here variance of number of trials on which we observe the first gold marble = (1 -0.3)/0.32 = 0.7/0.09 = 7.778

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