A spinner contains four sections: red, blue, green, and yellow. Joaquin spins the spinner twice. The set of outcomes is given as S = {RB, RG, RY, RR, BR, BG, BY, BB, GR, GB, GY, GG, YR, YB, YG, YY}. If the random variable is "yellow (Y)," which of the following is the correct probability distribution? A 2-column table has 3 rows. The first column is labeled Yellow: x with entries 0, 1, 2. The second column is labeled Probability with entries 0. 5625, 0. 375, 0. 625. A 2-column table has 3 rows. The first column is labeled Yellow: x with entries 0, 1, 2. The second column is labeled Probability with entries 0. 75, 0. 25, 0. A 2-column table has 3 rows. The first column is labeled Yellow: x with entries 0, 1, 2. The second column is labeled Probability with entries 0. 5, 0. 375, 0. 125. A 2-column table has 3 rows. The first column is labeled Yellow: x with entries 0, 1, 2. The second column is labeled Probability with entries 0. 5, 0. 25, 0. 25.

Probability distribution of a random variable is the collection of value and its probability pair for values of that considered random variable.

It might be a function, a table etc.

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

For this case, the experiment consists of a spinner with four color sections, and is spun twice.

The sample space is:

S = {RB, RG, RY, RR, BR, BG, BY, BB, GR, GB, GY, GG, YR, YB, YG, YY}

Let we take:

X = the number of times out of two spins the color yellow occurs in a pair of spin of the considered spinner.

Then, as there are only two spin going to happen, so the values X can take are:

X = 0 (no yellow color in both spin)
X = 1 (one of the spin ended up yellow, and other doesn't)

X = 2 (both the spins resulted in yellow color).

Let we take:

A = event that in first spin of the considered spinner, color yellow occurs.
B = event that in second spin of the considered spinner, color yellow occurs.

Then, as there are four colors equally likely, so we get:
P(A) = 1/4 (yellow color can occur in 1 way in one spin, and total number of colors that can occur = 4).

Similarly, we get P(B) = 1/4

Also, as first and second spins are independent of each other's result, so both A and B are independent events.

Getting the probabilities for each value of X:

Case 1: X = 0

[tex]P(X = 0) = P(A' \cap B') = P(A')P(B') = (1-P(A))(1-P(B)) \\P(X = 0) = \dfrac{3}{4} \times \dfrac{3}{4}\\\\P(X = 0) = \dfrac{9}{16}[/tex]

(where A' and B' are complement of A and B, and as (A and B) are independent, so as (A' and B') and (A' and B) and (A and B') )

Case 2: X = 1

So either first spin had yellow color, or second, so we get:

[tex]P(X = 1) = P((A' \cap B) \cup (A \cap B'))\\ P(X=1) = P(A' \cap B) + P(A \cap B') - P((A' \cap B) \cap (A \cap B'))\\P(X=1) = P(A')P(B) + P(A)P(B') -0\\P(X=1) = \dfrac{3}{4} \times \dfrac{1}{4} + \dfrac{1}{4} \times \dfrac{3}{4} - 0\\\\P(X = 1) = \dfrac{6}{16}[/tex]

Case 3: X = 2

Both event A and B occured this time, so we get:

[tex]P(X = 2) = P(A \cap B) = P(A)P(B) \\P(X = 0) = \dfrac{1}{4} \times \dfrac{1}{4}\\\\P(X = 0) = \dfrac{1}{16}[/tex]

Thus, the probability distribution of X is:

X=x P(X = x)

0 9/16 = 0.5625

1 6/16 = 0.375

2 1/16 = 0.0625

Learn more about probability here:

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