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A bag contains 6 red and 3 white balls. Four balls are drawn one by one and not replaced. The probability that they are alternately of different colors is: __________

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Final answer:

The probability of drawing four balls alternating in color from a bag containing 6 red and 3 white balls without replacement is calculated by considering both possible starting colors and summing their probabilities.

Step-by-step explanation:

The question involves calculating the probability of drawing four balls from a bag containing nine balls of different colors with alternating colors, without replacement. In this case, there are 6 red and 3 white balls. The calculation involves determining the probability for two different scenarios: starting with a red ball and starting with a white ball, as these are independent events with their probabilities.

Calculation Steps:

  1. Determine the total possible ways to draw four balls without replacement.
  2. Calculate the probability for the sequence starting with a red ball: Red-White-Red-White.
  3. Calculate the probability for the sequence starting with a white ball: White-Red-White-Red.
  4. Add the two probabilities together for the final answer.

Let's consider both sequences:

For Red-White-Red-White (RWRW):

The first ball has a 6/9 chance of being red, the second has 3/8 (since one red has been taken out and not replaced) to be white, the third has 5/7 of being red, and the fourth has 2/6 or 1/3 of being white.

The probability in this case is (6/9)*(3/8)*(5/7)*(1/3).

For White-Red-White-Red (WRWR):

The first ball has a 3/9 chance of being white, the second has 6/8 or 3/4 of being red, the third has 2/7 of being white, and the fourth has 5/6 of being red.

The probability in this case is (3/9)*(3/4)*(2/7)*(5/6).

The combined probability will be the sum of these two individual probabilities.

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