Question

suppose that each circle is equally likely to be selected. one circle is selected at random. determine the probability indicated.

\\(p(-\mid \text{yellow circle obtained})\\)

Explanation:

Identify the sample space and given condition

The set of circles shown in the top right contains 7 circles with signs:

• Yellow circle with \(+\)
• Yellow circle with \(-\)
• Yellow circle with \(-\)
• Red circle with \(-\)
• Yellow circle with \(+\)
• Blue circle with \(+\)

Let's count the total number of yellow circles, which represents the given event \(E_1\) ("yellow circle obtained"):

• There are 4 yellow circles in total:

1. Yellow with \(+\)
2. Yellow with \(-\)
3. Yellow with \(-\)
4. Yellow with \(+\)

Thus, \(n(E_1) = 4\).

Count the favorable outcomes under the condition

We want to find the probability of obtaining a minus sign given that a yellow circle is obtained: \(P(-\mid \text{yellow circle obtained})\).
The event \(E_2\) is "obtaining a minus sign".
The intersection event "\(E_1 \text{ and } E_2\)" represents obtaining a yellow circle with a minus sign.
Looking at our yellow circles:

• Yellow with \(-\) (2 circles)

Thus, \(n(E_1 \text{ and } E_2) = 2\).

Calculate the conditional probability

Using the conditional probability formula:

$$ P(E_2 \mid E_1) = \frac{n(E_1 \text{ and } E_2)}{n(E_1)} = \frac{2}{4} = \frac{1}{2} $$

Answer:

Suppose that each circle is equally likely to be selected. One circle is selected at random. Determine the probability indicated.

\(P(-\mid \text{yellow circle obtained}) =\) <blank>\(\frac{1}{2}\)</blank>