Answered

A spinner is divided into two equal parts, one red and one blue. The set of possible outcomes when the spinner is spun twice is $S =\{ RR , RB , BR , BB \}$. Let $X$ represent the number of times blue occurs. Which of the following is the probability distribution, $P_{ X }( x )$?



A.


$x$		$P_X(x)$

0		0.25

1		0.5

2		0.25






B.


$x$		$P_X(x)$

0		0.33

1		0.33

2		0.33






C.


$x$		$P_X(x)$

0		0.5

1		0.5

2		0






D.


$x$		$P_X(x)$

0		0

Verified Answer :

  • Each outcome of the two spins has a probability of \(0.5 \times 0.5 = 0.25\).
  • Mapping the number of blues (\(X\)) to each outcome gives: \(X(RR)=0\), \(X(RB)=1\), \(X(BR)=1\), and \(X(BB)=2\).
  • The probabilities for each \(X\) are \(P(X=0)=0.25\), \(P(X=1)=0.25+0.25=0.5\), and \(P(X=2)=0.25\).
  • The correct distribution is the first table: \(\boxed{\text{First Table}}\).

Explanation

  1. Analyze the experiment and sample space

The spinner is divided into two equal parts, one red and one blue, implying that the probability of landing on either color is \(\frac{1}{2} = 0.5\). When the spinner is spun twice, the sample space of possible outcomes is \(S = \{RR, RB, BR, BB\}\). Since each spin is independent, the probability of any specific sequence is the product of the individual probabilities:
\[P(RR) = 0.5 \times 0.5 = 0.25\]
\[P(RB) = 0.5 \times 0.5 = 0.25\]
\[P(BR) = 0.5 \times 0.5 = 0.25\]
\[P(BB) = 0.5 \times 0.5 = 0.25\]

  1. Define the random variable X

The random variable \(X\) is defined as the number of times blue occurs in the two spins. Let's map each outcome in the sample space \(S\) to its corresponding value of \(X\):

  • For \(RR\), the number of blues is \(0\), so \(X(RR) = 0\).
  • For \(RB\), the number of blues is \(1\), so \(X(RB) = 1\).
  • For \(BR\), the number of blues is \(1\), so \(X(BR) = 1\).
  • For \(BB\), the number of blues is \(2\), so \(X(BB) = 2\).
  1. Determine the probability distribution table

Now we calculate the probability distribution \(P_X(x)\) for each possible value of \(x \in \{0, 1, 2\}\):

  • For \(x=0\): The only outcome is \(RR\). Therefore, \(P_X(0) = P(RR) = 0.25\).
  • For \(x=1\): The outcomes are \(RB\) and \(BR\). Therefore, \(P_X(1) = P(RB) + P(BR) = 0.25 + 0.25 = 0.5\).
  • For \(x=2\): The only outcome is \(BB\). Therefore, \(P_X(2) = P(BB) = 0.25\).

Summing these probabilities confirms the distribution is valid, as \(0.25 + 0.5 + 0.25 = 1\). Comparing these results with the given options, the first table is the correct one.

Examples

Probability distribution tables like this are fundamental in fields like statistics and quality control. For example, if a machine produces components that are either "defective" or "functional" with a 50% chance, this model helps determine the likelihood of encountering 0, 1, or 2 defective parts in a batch of 2, allowing engineers to estimate failure rates accurately. The calculation \(P(X=1) = P(RB) + P(BR) = 0.25 + 0.25 = 0.5\) is a practical application of the sum rule for independent events in a sample space.

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