Question

brendan has two bags of marbles. the two bags look the same, and each contains five marbles. in one bag, there are one white marble and four red marbles. in the other bag, there are four white marbles and one red marble. caylin randomly selects one bag of marbles. she cannot see the marbles in the bag, and only knows the marbles she has drawn. she then randomly draws a marble from the bag she selected. she sees that it is a red marble. based on this observation, what is the probability that the bag she selected is the one that contains one white marble and four red marbles? 1. before drawing any marbles, explain why the probability that caylin selected the bag that contains one white marble and four red marbles is \\(\frac{1}{2}\\).

Explanation:

Part 1: Explanation for Probability Before Drawing

There are two identical - looking bags, and Caylin randomly selects one. Since there are 2 possible bags and each bag has an equal chance of being selected (because the selection is random and the bags look the same), the probability of selecting the bag with 1 white and 4 red marbles is the number of favorable outcomes (1 bag) divided by the total number of possible outcomes (2 bags), so $P=\frac{1}{2}$.

Step 1: Identify the values of $P(A)$, $P(

eg A)$, $P(B|A)$ and $P(B|
eg A)$
We have $P(A)=\frac{1}{2}$, $P(
eg A)=\frac{1}{2}$, $P(B|A)=\frac{4}{5}$, $P(B|
eg A)=\frac{1}{5}$.

Step 2: Substitute the values into Bayes' formula

First, calculate the numerator: $P(B|A)P(A)=\frac{4}{5}\times\frac{1}{2}=\frac{4}{10}$
Then, calculate the denominator: $P(B|A)P(A)+P(B|
eg A)P(
eg A)=\frac{4}{5}\times\frac{1}{2}+\frac{1}{5}\times\frac{1}{2}=\frac{4 + 1}{10}=\frac{5}{10}$

Step 3: Calculate $P(A|B)$

$P(A|B)=\frac{\frac{4}{10}}{\frac{5}{10}}=\frac{4}{5}$

Answer:

Because there are 2 identical - looking bags, and Caylin randomly selects one. The probability of selecting a particular bag (the one with 1 white and 4 red marbles) is the number of that bag (1) divided by the total number of bags (2), so the probability is $\frac{1}{2}$.

Part 2: Probability After Drawing a Red Marble (Using Bayes' Theorem)

We will use Bayes' theorem. Let's define:

• Let $A$ be the event that the selected bag is the one with 1 white and 4 red marbles.
• Let $B$ be the event that a red marble is drawn.

We know that:

• $P(A)=\frac{1}{2}$ (probability of selecting the first bag before drawing)
• $P(

eg A)=\frac{1}{2}$ (probability of selecting the second bag before drawing)

• $P(B|A)$: probability of drawing a red marble given that we selected the first bag. In the first bag, there are 5 marbles, 4 of which are red. So $P(B|A)=\frac{4}{5}$.
• $P(B|

eg A)$: probability of drawing a red marble given that we selected the second bag. In the second bag, there are 5 marbles, 1 of which is red. So $P(B|
eg A)=\frac{1}{5}$.

By Bayes' theorem, $P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|
eg A)P(
eg A)}$