24.4k views
2 votes
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: __________

1 Answer

2 votes

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.

User PineapplePie
by
8.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.