Question

10 red marbles and 10 blue marbles are placed into a bag. Alex mixes up the bag and randomly selects a marble. He continues to do so, replacing the marble after each selection, until a red marble is selected.
a. What is the probability that the first time that a red marble is pulled is on Alex’s 6th try?
b. On average, how many marbles will Alex have to pull in order to get a red marble? (Hint: use math expectation)

Answer 1

Answer on Question #55335 – Math – Statistics and Probability

Question

10 red marbles and 10 blue marbles are placed into a bag. Alex mixes up the bag and randomly selects a marble. He continues to do so, replacing the marble after each selection, until a red marble is selected.

a. What is the probability that the first time that a red marble is pulled is on Alex’s 6th try?

b. On average, how many marbles will Alex have to pull in order to get a red marble? (Hint: use math expectation).

Solution

a. Plainly, there are $10 + 10 = 20$ marbles in the bag. Using the classical definition of probability, the probability of selecting a red marble is given by

$$p = \frac{10}{20} = 0.5.$$

Similarly, probability of selecting a blue marble is

$$q = \frac{10}{20} = 0.5 = 1 - p.$$

If marbles are replaced, then probability remains the same for all these experiments. We are asked what are the odds of selecting a red marble for the first time on the $6^{\text{th}}$ try. It means that we are asked to determine a probability of the case we will call Q (Alex selects 5 blue marbles on five first tries and a red one on the 6-th try).

Events of marble selection are independent, therefore

$$\begin{array}{l} P(Q) = P(1^{st} = \text{blue and } 2^{nd} = \text{blue and ... and } 5^{th} = \text{blue and } 6^{th} = \text{red}) = \\ = P(1^{st} = \text{blue}) \cdot P(2^{nd} = \text{blue}) \cdot \ldots \cdot P(5^{th} = \text{blue}) \cdot P(6^{th} = \text{red}) = \\ = q \cdot q \cdot \ldots \cdot q \cdot p = q^5 \cdot p = 0.5^5 \cdot 0.5 \approx 0.016 \text{ or } P(Q) = 1.6\% \end{array}$$

Answer: 0.016.

Solution

b. Average number of pulls (denoted by $E(X)$) is a mathematical expectation of number of pulls. If Alex pulls a red marble on the $1^{\text{st}}$ try, then the number of such pulls will be

$$x_1 = 1.$$

Probability of this event is just

$$p_1 = p = 0.5.$$

Now, if it happens on the $2^{\text{nd}}$ try, the number is

$$x_2 = 2,$$

and the probability is

$$p_2 = q \cdot p = p^2,$$

because this event is the following:

Alex selecting a blue marble on the first try (with probability q) and then selecting a red one (with probability p). Now, for 3-rd try the number of pulls is

$$x_3 = 3,$$

and the probability is

$$p _ {3} = q \cdot q \cdot p = p ^ {3}.$$

Observing the pattern, we get formulae for the n-th try:

number of pulls is

$$x _ {n} = n$$

and probability is

$$p _ {n} = q \cdot \ldots \cdot q \cdot p = q ^ {n - 1} \cdot p = p ^ {n}.$$

Then, by definition of the math expectation, we have

$$\mathrm {E} (\mathrm {X}) = \sum_ {k = 1} ^ {\infty} x _ {k} p _ {k} = x _ {1} p _ {1} + x _ {2} p _ {2} + x _ {3} p _ {3} + \dots = \sum_ {k = 1} ^ {\infty} k p ^ {k} = \frac {p}{(1 - p) ^ {2}} = \frac {0 . 5}{(1 - 0 . 5) ^ {2}} = \frac {1}{0 . 5} = 2.$$

Answer: 2.

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