Answered

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.


Yellow: X		Probability: $P_x(x)$

0		0.5625

1		0.375

2		0.0625






B.


Yellow: X		Probability: $P_{X}(x)$

0		0.75

1		0.25

2		0






C.


Yellow: X		Probability: $P_{X}(x)$

0		0.5

1		0.375

2		0.125

Verified Answer :

  • We analyze the sample space \(S\) of 16 equally likely outcomes to count the occurrences of yellow (Y).
  • Outcomes with zero Y: 9 occurrences, \(P(X=0) = \frac{9}{16} = 0.5625\).
  • Outcomes with one Y: 6 occurrences, \(P(X=1) = \frac{6}{16} = 0.375\).
  • Outcomes with two Y: 1 occurrence, \(P(X=2) = \frac{1}{16} = 0.0625\).
  • The correct probability distribution is:

\(\boxed{\text{First Table}}\)

Explanation

  1. Understand the problem and the sample space

To determine the correct probability distribution for the random variable \(X\), representing the "number of times yellow is spun in two spins," we first examine the total sample space provided:
\(S = \{RB, RG, RY, RR, BR, BG, BY, BB, GR, GB, GY, GG, YR, YB, YG, YY\}\).
There are 16 total possible outcomes. Since each outcome is assumed to be equally likely, the probability of any specific outcome is \(\frac{1}{16} = 0.0625\).

  1. Calculate the probabilities for X = 0, 1, and 2

We count how many times yellow ('Y') appears in each of the 16 outcomes:

  1. Outcomes with 0 yellows: {RB, RG, RR, BR, BG, BB, GR, GB, GG}. There are 9 such outcomes.

Probability \(P(X=0) = \frac{9}{16} = 0.5625\).

  1. Outcomes with 1 yellow: {RY, BY, GY, YR, YB, YG}. There are 6 such outcomes.

Probability \(P(X=1) = \frac{6}{16} = 0.375\).

  1. Outcomes with 2 yellows: {YY}. There is 1 such outcome.

Probability \(P(X=2) = \frac{1}{16} = 0.0625\).

  1. Determine the correct table

Comparing our results to the given options:

  • \(P(X=0) = 0.5625\)
  • \(P(X=1) = 0.375\)
  • \(P(X=2) = 0.0625\)

This distribution corresponds to the first table provided in the question.

Examples

Probability distributions like this are essential in fields such as game design and statistics. For example, if you are designing a board game that uses a spinner, you can calculate the likelihood of different outcomes to ensure the game is balanced and challenging for the players. By understanding that spinning yellow zero times is much more likely (\(56.25\%\)) than spinning it twice (\(6.25\%\)), you can adjust the rules to make certain rewards or penalties rarer and thus more valuable.

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