Question

A jar contains 11 red marbles, 12 blue marbles, and 6 white marbles. Four marbles from this jar are selected, with each marble being replaced after each selection.

What is the probability that the first white marble chosen is on the 4th selection?

0.1032
0.3957
0.5011
0.6043

Answer 1

Answer:

Approximately $0.1032$.

Step-by-step explanation:

The marbles are returned to the jar ("replaced") after each draw, meaning that the content of the jar stays the same. Thus, the probability distribution of each draw would be independent and identical.

Let $X_{1}$, $X_{2}$, $X_{3}$, and $X_{4}$ denote whether each draw is a white marble, with $X = 1$ if the draw is a white marble and $X = 0$ otherwise.

Drawing the first white marble on the $4$th selection means:

• choosing a marble that is not a white marble at the first $3$ selections, $(X_{1} = 0) \cap (X_{2} = 0) \cap (X_{3} = 0)$, and
• choose a white marble at the $4$th selection, $(X_{4} = 1)$.

The corresponding probability would be:

$P(X_{1} = 0 \cap X_{2} = 0 \cap X_{3} = 0 \cap X_{4} = 1)$.

Since the draws are independent from one another:

$\begin{aligned} & P(X_{1} = 0 \cap X_{2} = 0 \cap X_{3} = 0 \cap X_{4} = 1) \\ =\; & P(X_{1} = 0)\, P(X_{2} = 0)\, P(X_{3} = 0) \, P(X_{4} = 1)\end{aligned}$.

There are a total of $11 + 12 + 6 = 29$ marbles in this jar.

With $6$ out of these $29$ marbles being white marbles, the probability of choosing a white marble at each draw would all be $P(X = 1) = (6 / 29)$. The probability of not choosing a white marble at each draw would then be $P(X = 0) = (1 - (6/29))$. Therefore:

$\begin{aligned} & P(X_{1} = 0 \cap X_{2} = 0 \cap X_{3} = 0 \cap X_{4} = 1) \\ =\; & P(X_{1} = 0)\, P(X_{2} = 0)\, P(X_{3} = 0) \, P(X_{4} = 1) \\ =\; & \left(1 - \frac{6}{29}\right)\, \left(1 - \frac{6}{29}\right)\, \left(1 - \frac{6}{29}\right)\, \left(\frac{6}{29}\right) \\ =\; & \frac{73\, 002}{707\, 281}\\ \approx \; & 0.1032\end{aligned}$.