Question

(20) 4. Suppose I have a bag of 16 marbles and 11 are blue. The rest are other colors. (1) C1. What type of probability distribution is this if I want to know the probability of choosing three blue marbles in a row if the marble is replaced? (3) C2. What is the probability of choosing three blue marbles in a row if the marble is replaced? Show how to solve this. (1) C3. How many blue marbles should I expect to choose in a row? (1) D1. What type of probability distribution is this if I want to know the probability of choosing 6 other colored marbles in a row before I choose a blue marble if the marble is replaced each time I choose it? (3) D2. What is the probability of choosing 6 other colored marbles in a row if the marble is replaced? Show how to solve this. (1) D3. How many blue marbles should I expect to choose in a row?

Answer 1

c1) Given that the marble is replaced everytime, the probability of getting a blue marble remains same in each case, therefore the distribution here for the number of blue marbles drawn in 3 draws is modelled here as:

$X \sim Bin(n = 3, p= 11/16)$

That is a binomial distribution here

c2) The probability here is computed here as:

Therefore 0.3250 is the required probability here.

c3) The number of marbles expected to be chosen in a row is computed here as:

P(Y = 0) = (5/16)
P(Y = 1) = (11/16)*(5/16)
P(Y = 2) = (11/16)²*(5/16)

and so on...

Therefore the expected value here is computed as:
E(X) = 1P(X= 1) + 2P(X = 2) + 3P(X = 3) + ...... Infinity

E(Y) = (11/16)(5/16) + 2*(11/16)²*(5/16) +  3*(11/16)³*(5/16) + ... infinity

(11/16)E(X) = (11/16)²(5/16) + 2*(11/16)³*(5/16) +  3*(11/16)⁴*(5/16) + ... infinity

Subtracting the last equation from the second last one, we have here:

(5/16)E(X) = (11/16)(5/16) + (11/16)²(5/16) + (11/16)³*(5/16) +.... infinity

E(X) = (11/16) + (11/16)² + (11/16)³ + ..... + infinity

$E(X) = \frac{(11/16)}{1 - (11/16)} = \frac{11}{5} = 2.2$

Therefore 2.2 is the expected number of blue marbles here.

d1) Again, the probability distribution here would be a binomial distribution but with parameters n = 6 and p = 5/16

d2) The probability here is computed as:

= (5/16)⁶ = 0.000931

Therefore 0.000931 is the required probability here.

d3) The expected number of marbles in a row here is computed as:
= 5/11 = 0.4545

Therefore 0.4545 marbles is the correct answer here.