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An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable ?

User Damary
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Final answer:

The random variable X can take on the values of 0, 1, or 2, which represent the number of black balls drawn from an urn. It is indeed a random variable because its value is not known prior to the drawing and depends on a random process.

Step-by-step explanation:

The values that the random variable X, representing the number of black balls drawn from an urn, can take are 0, 1, or 2. These values arise from the possible outcomes when drawing two balls without replacement from an urn that contains 5 red and 2 black balls. If no black ball is drawn, then X = 0; if one black ball is drawn, then X = 1; and if both balls drawn are black, then X = 2.

The number of black balls, X, can take on either 0 or 1 as values. If both balls drawn are red, then X = 0. If one of the balls drawn is black, then X = 1. Therefore, the possible values of X are 0 and 1.

Yes, X is a random variable. A random variable is a numerical value that represents the outcomes of a random experiment. In this case, X represents the number of black balls drawn, which can vary depending on the outcome of the experiment.

Yes, X is a random variable because it is a numerical quantity that arises from a random process of drawing balls from the urn. The value of X is determined by the outcome of this process, which cannot be predicted with certainty ahead of time, making it a variable that can take on different values according to the random draw.

User Idle Sign
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